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Naddika [18.5K]
3 years ago
10

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.​

Engineering
1 answer:
Novosadov [1.4K]3 years ago
5 0

Answer:

i) SF: v(x) = \frac{(w_0* x )^2}{2L}

ii) BM : = \frac{(w_0*x)^3}{6L}

Explanation:

Let's take,

\frac{y}{w_0} = \frac{x}{L}

Making y the subject of formula, we have :

y = \frac{x}{L} * w_0

For shear force (SF), we have:

This is the area of the diagram.

v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0

= \frac{(w_0* x )^2}{2L}

The shear force equation =

v(x) = \frac{(w_0* x )^2}{2L}

For bending moment (BM):

BM = v(x) * \frac{x}{3}

= \frac{(w_0* x )^2}{2L}  * \frac{x}{3}

= \frac{(w_0*x)^3}{6L}

The bending moment equation =

= \frac{(w_0*x)^3}{6L}

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Assoli18 [71]

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

pressure at the bottom is =55\times 25

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1 ft = 12 inch

pressure at bottom =\frac{1375}{12^2}

                                = 9.55 psi

so, answer will be option (d) which is 9.55 psi

3 0
3 years ago
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough
jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection  

Explanation:

We are given:

Plane strain fracture toughness K = 98.9 MPa \sqrt{m}

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

oc = \frac{K}{Y \sqrt{pi * a}}

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

a= \frac{1}{pi} * [\frac{k}{y*a}]^2

Substituting figures in the expression above, we have:

= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection  

3 0
3 years ago
Read 2 more answers
(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1
olya-2409 [2.1K]

Answer:

The answer is "4.35 \times 10^{-3}\  mm and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point  

y =  yo + \frac{k}{\sqrt{x}}

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.  

Replacing the equation for each condition:  

y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k

People can get yo = 275 MPa with both equations and k= 15.5 Mpa mm^{\frac{1}{2}}.

To substitute the answer,  

310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\  mm

In point b:

The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

equation is:

75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\

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lilavasa [31]

Answer:

See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void printNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

printNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

printNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

printNumPattern(num1, num2);

}

}

See attachment for sample output

3 0
3 years ago
Read 2 more answers
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antoniya [11.8K]

Answer:

Velocity boundary layer thickness over an internal or internal fixed flat plate is  normal to surface distance  required for the fluid velocity to attain the 99% of the free stream velocity value.

Explanation:

The affirmation is the current definition of velocity boundary layer thickness.

4 0
2 years ago
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