) You are using a load cell to measure the applied load to a test in the Civil Engineering Structures Lab. What should you do if the measured load does not return to zero? How should you troubleshoot this to determine if this is a load cell or a mechanical problem?

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Based on the available information, the the estimated towing power required is:

(a) For the parallel orientation: 655 kW

(b) For the normal orientation: 4,116.75 kW

Given:

- Diameter of the cylinder: 1.5 m

- Length of the cylinder: 22 m

- Towing speed (U): 5 m/s

- Water temperature: 20°C

(a) Cylinder parallel to the tow direction:

Step 1: Calculate the drag force for the parallel orientation.

The drag force for a submerged cylinder in parallel orientation is given by the formula:

F_D = 0.5 × ρ × C_D × A × U^2

Where:

- ρ is the density of fresh water at 20°C, which is approximately 998 kg/m³.

- C_D is the drag coefficient for a cylinder in parallel orientation, which is approximately 0.82.

- A is the cross-sectional area of the cylinder, which is π × D × L = π × 1.5 m × 22 m = 103.87 m².

Calculating the drag force:

F_D = 0.5 × 998 kg/m³ × 0.82 × 103.87 m² × (5 m/s)² = 131,000 N

Step 2: Calculate the towing power for the parallel orientation.

Towing power = Drag force × Towing speed

Towing power = 131,000 N × 5 m/s = 655,000 W = 655 kW

(b) Cylinder normal to the tow direction:

Step 3: Calculate the drag force for the normal orientation.

The drag force for a submerged cylinder in normal orientation is given by the formula:

F_D = 0.5 × ρ × C_D × A × U^2

Where:

- C_D is the drag coefficient for a cylinder in normal orientation, which is approximately 1.2.

- A is the cross-sectional area of the cylinder, which is π × D × D/4 = π × 1.5 m × 1.5 m/4 = 1.77 m².

Calculating the drag force:

F_D = 0.5 × 998 kg/m³ × 1.2 × 1.77 m² × (5 m/s)² = 823,350 N

Step 4: Calculate the towing power for the normal orientation.

Towing power = Drag force × Towing speed

Towing power = 823,350 N × 5 m/s = 4,116,750 W = 4,116.75 kW

Therefore, the estimated towing power required is:

(a) For the parallel orientation: 655 kW

(b) For the normal orientation: 4,116.75 kW

Answer:

it will take for the casting to solidify 2.55 min

Explanation:

given data

mold constant = 4 min/cm²

length = 35 cm

width = 10 cm

thickness = 15 mm

solution

we use here Chvorinov's Rule that is

Chvorinov's Rule = mold constant × [tex](\frac{V}{A})^{1.9}[/tex]   ..............1

put here value

Chvorinov's Rule = 4 × [tex](\frac{600}{760})^{1.9}[/tex]  

Chvorinov's Rule = 2.55 min/in

so heer unit flow become [tex]min/in^{1.9}[/tex]  

1225

Explanation:

This segment helps initialize sum as 0. The for loop is used to increment with every execution and it is added to the sum. The loop runs 49 times and every time the count is added to the sum. In short it is the sum of first 49 natural numbers i.e 1+2+3+......+49.

Answer:

The solution to the given problem is provided below.

Explanation:

a.) myBike.height = 26;                                                 Not Legal statement

b.) myBike.model = “Cyclone”:                                    Legal statement

c.) myBike.wheels = 3;                                                  Legal statement

d.) myBike.model = 108;                                               Not legal statement

e.) Bicycle.height = 24;                                                  Not Legal statement

f.) Bicycle.model = “Hurricane”;                                 Not legal statement

g.) Bicycle.int = 3;                                                           Not Legal statement

h.) Bicycle.model = 108;                                                Not Legal Statement

i.) Bicycle.wheels = 2;                                                    Legal Statement

j.) Bicycle yourBike = myBike;                                      Legal Statement

Answer:

Explanation: see the pictures attached

Answer:

Heat lost from the cylindrical pipe is given by the formula,

[tex]d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }[/tex]

Temperature distribution inside the cylinder is given by,

[tex]\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}[/tex]    

Explanation:

Inner radius = [tex]R_{1}[/tex]  

Outer radius = [tex]R_{2}[/tex]

Constant thermal conductivity = K

Inner temperature = [tex]T_{1}[/tex]

Outer temperature = [tex]T_{2}[/tex]

Length of the pipe = L

Heat lost from the cylindrical pipe is given by the formula,

[tex]d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }[/tex]------------ (1)

Temperature distribution inside the cylinder is given by,

[tex]\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}[/tex]     ------------ (2)

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

[tex]F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}[/tex]

The bouyancy force is:

[tex]F_{B} =V*p_{w} *g[/tex]

The weight is:

[tex]W=V*p_{c} *g[/tex]

Laminar flow:

[tex]v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s[/tex]

Reynold number:

[tex]Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1[/tex]

Not flow region

For Newton flow region:

[tex]v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s[/tex]

[tex]Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4[/tex]

[tex]Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31[/tex]

[tex]Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99[/tex]

[tex]Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K[/tex]

[tex]\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s[/tex]

[tex]e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m[/tex]

Answer:

The answer is True.

Explanation:

SED is a command in UNIX for stream editors that parses and transforms text, using a simple, compact programming language. So the answer is TRUE that sed is a multipurpose tool that combines the work of several filters and performs noninteractive operations on a data stream. sed has a host of features that allow you ti select lines and run instructions on them.

To calculate the required cross-sectional area of copper wires needed to connect the DC supply source to the mine lights with less than 5% power loss, calculate the permissible power loss, and then use the relationship between power loss, current, and wire resistance. Considering the resistivity of copper and the round trip length of the wire, the formula A = ρL/(δP/I²) determines the necessary wire thickness.

To determine the required cross-sectional area of copper wires for the lighting equipment, we need to ensure the power loss (δP) is no more than 5% of the lights' power usage (P), which is 5 kW (5000 W). As the power loss in wires is given by δP = I²R, where I is the current and R is the resistance of the wire, we must first calculate the current using P = IV, giving I = P/V = 5000W/120V = 41.67 A. The power loss allowed is thus 5% of 5000 W, equating to 250 W. Given δP, we can find R using δP = I²R, which gives R = δP/I².

The resistance of a copper wire is also given by R = ρL/A, where ρ is the resistivity of copper (1.68 x 10^-8 Ωm), L is the length of the wire (200 meters round trip), and A is the cross-sectional area we need to find. Equating the two expressions for R and solving for A gives A = ρL/(δP/I²). Substituting the given and calculated values yields the required cross-sectional area. Finally, as resistance depends on the entire length of the circuit, remember to double the distance to account for both the outgoing and return paths.

Answer:

5.24m

Explanation:

Data given

force, F= 11,100N

safety factor,N=2

yield strength, =1030MPa

To determine the minimum diameter, we first determine the working strength which is expressed as

[tex]working strength=\frac{yield strength}{safty factor}\\ Working strength =1030/2=515MPa\\[/tex]

Since also the working strength is define as the ration of the force to the area, we have

[tex]515=\frac{11100}{A}\\ A=21.55m^{2}[/tex]

hence the required diameter is given as

[tex]A=\pi d^2/4\\d=\sqrt{\frac{4*21.55 }{\pi }} \\d=5.24m[/tex]

Determining the radius of a brass cylinder based on its elongation under tensile load involves understanding stress, strain, and the material's properties. Without the elastic modulus of brass or additional details, an exact calculation can't be provided. Theoretically, one uses the relationship between stress, strain, and Young's modulus, and the formula for the area of a circle to find the radius.

To determine the radius of a cylindrical specimen of brass that must elongate a specific amount under a given tensile load, one must approach the problem by considering the relationship between stress, strain, and the elastic modulus of the material. Given that the specimen is cylindrical, the cross-sectional area is crucial in these calculations, which is directly related to the radius of the cylinder.

The formula for stress (σ) is defined as the force (F) divided by the area (A), σ = F/A. Strain (ε) is the deformation (change in length) divided by the original length (L), ε = ΔL/L. However, without the elastic modulus of brass or a direct way to calculate the cross-sectional area from the provided data, finding the exact radius requires assuming or knowing additional properties of the material.

Without these specifics, a more detailed calculation cannot be accurately provided. In practice, one would use the known properties of brass and the relationship between stress, strain, and Young's modulus (Y) of the material (Y = stress/strain) to find the required dimensions. Typically, this involves rearranging the formulas to solve for the radius, given that the area (A) can be expressed in terms of the radius (r) for a circle (A = πr²).

Explanation:

Please refer to the attached image

Python Code:

Please refer to the attached image

Output:

Please refer to the attached image

Following are the program to print the first 128 ASCII values:

#include <iostream>//header file

using namespace std;

int main()//main method

{

int i=1;//defining integer variable

for(i=1;i<=128;i++) //defining loop that prints the first 128 ASCII values

{

   char x=(char)i;//defining character variable that converts integer value into ASCII code value

   printf("%d=%c\n",i ,x);//print converted ASCII code value

}

   return 0;

}

Program Explanation:

Output:

Please find the attached file.

Find out more information about the ASCII values here:

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Answer:

B

Explanation:

Torsion is application of torque to a shaft to turn it about its longitudinal axis. When torque is applied to a shaft the circle remains unchanged in a circular state, its cross section does not warp but remains flat with a straight radial lines but its longitudinal lines changes into an helix intersecting the circular shaft

Answer:

A. 0.020

B. 0.018

Explanation: check the attached file

Answer:

a. n=0.020  b. n=0.018

Explanation:

a.

Case 1: To line the sides only

n=(Σn₁¹°⁵P₁)^2/3/P^2/3

n = equivalent roughness

n₁=corresponding roughness coefficients

P=length

At the bed: n₁=0.025 and P₁=5m

At the sides: n₂=0.012 and P₂= 2*1.1*√1+1.5²=2.2*1.8=3.96m

P = P₁+P₂=8.96m

Equivalent roughness, n = [5*(0.025)^1.5+3.96*(0.012)^1.5]^2/3/(8.96)^2/3

n= [(5*0.00395)+(3.96*0.0013)]^2/3/4.317

n=0.0249^2/3/4.317

n=0.0842/4.317

n=0.0195

n=0.020

b.

Case 2: To line the bed only

P₁=5m   n₁=0.012

P₂=3.96 n₂=0.025

P=8.96

Equivalent roughness n= [5*(0.012)^1.5+3.96*(0.025)^1.5]^2/3/(8.96)^2/3

n=[(5*0.0013)+(3.96*0.00395)]^2/3/4.317

n=0.0221^2/3/4.317

n=0.078/4.317

n=0.018

Answer:

[tex]q_{out} = 9.25\,\frac{kJ}{kg}[/tex]

Explanation:

First, it is required to find the absolute humidity of air at initial state:

[tex]\omega = \frac{0.622\cdot \phi \cdot P_{g}}{P-\phi \cdot P_{g}}[/tex]

The saturation pressure at [tex]T = 27^{\textdegree}C[/tex] is:

[tex]P_{g} = 3.601\,kPa[/tex]

Then,

[tex]\omega = \frac{0.622\cdot (0.5)\cdot (3.601\,kPa)}{101.325\,kPa-(0.5)\cdot (3.601\,kPa)}[/tex]

[tex]\omega = 0.0113\,\frac{kg\,H_{2}O}{kg\,air}[/tex]

A simple cooling process implies a cooling process with constant absolute humidity. The specific entalphies for humid air are:

Initial state:

[tex]h_{1} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (27^{\textdegree}C)+(0.0113)\cdot (2551.96\,\frac{kJ}{kg} )[/tex]

[tex]h_{1} = 55.972\,\frac{kJ}{kg}[/tex]

Final state:

[tex]h_{2} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (18^{\textdegree}C)+(0.0113)\cdot (2533.76\,\frac{kJ}{kg} )[/tex]

[tex]h_{2} = 46.722\,\frac{kJ}{kg}[/tex]

The specific energy that is removed is:

[tex]q_{out}= h_{1} - h_{2}[/tex]

[tex]q_{out} = 9.25\,\frac{kJ}{kg}[/tex]

Answer:

The most common approach to accessing an array is to use a for loop:

var mammals = new Array("cat","dog","human","whale","seal");

var animalString = "";

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

animalString += mammals[i] + " ";

}

alert(animalString);

Discussion

A for loop can be used to access every element of an array. The array begins at zero, and the array property length is used to set the loop end.

Though support for both indexOf and lastIndexOf has existed in browsers for some time, it’s only been formalized with the release of ECMAScript 5. Both methods take a search value, which is then compared to every element in the array. If the value is found, both return an index representing the array element. If the value is not found, –1 is returned. The indexOf method returns the first one found, the lastIndexOf returns the last one found:

var animals = new Array("dog","cat","seal","walrus","lion", "cat");

alert(animals.indexOf("cat")); // prints 1

alert(animals.lastIndexOf("cat")); // prints 5

Both methods can take a starting index, setting where the search is going to start:

var animals = new Array("dog","cat","seal","walrus","lion", "cat");

alert(animals.indexOf("cat",2)); // prints 5

alert(animals.lastIndexOf("cat",4)); /

Answer:

animals = ["Dog", "Lion", "Goat", "Zebra", "Cat"]

           for animal in animals:

                       print(animal)

x = animals.insert(5, "Lizard")

y = animals.insert(6, "Bat")

z = sorted(animals)

print(z)

Explanation:

The question can be solved using various back-end coding language like python, java, JavaScript etc. But I will be writing the code with python.

The first question said we should create an array or list of five animals.

animals = ["Dog", "Lion", "Goat", "Zebra", "Cat"]  → I created the five animals in a list and stored them in the variable animals.

for animal in animals:  I used a for loop to iterate through the list print(animal)  I used print statement to display the values stored in the array or list.

x = animals.insert(5, "Lizard")  I added an animal, lizard to the end of the array

y = animals.insert(6, "Bat")  I added another animal , Bat to the end of the array.

z = sorted(animals)  I sorted the array according to alphabetical number.

print(z) I displayed the sorted array .

Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.

Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.  

One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.  

Answer:

The answer is in the attachment.

Explanation:

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

Please find the attached file for the calculation of the 4 environment solutions:

Given:

[tex]f_c = 900\ MHz\\\\ h_t = 20\ m\\\\ h_r = 5\ m\\\\ d = 100\ m\\[/tex]

To find:

environments=?

Solution:

Please find the attached file.

Learn more about the Qualitative:

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Answer:

Explanation:

The detailed steps and calculations is as shown in the attachment.

where s = flow rate of incoming stream

Cs = concentration of pollutant in stream

Qf = flow rate from factory

Cp = concentration of pollutant from factory

Qo = flow rate out of lake

Co = concentration of pollutant in the lake

V = volume of lake

k = reaction rate coefficient

Using the maximum stress limit and elongation criteria, the required diameter of the aluminum rod to withstand a 3-kN axial load without exceeding a 1 mm stretch and keeping the normal stress below 40 MPa is calculated to be approximately 9.8 mm.

Determining the Required Diameter of an Aluminum Rod

To ensure a 1.5-m-long aluminum rod does not stretch more than 1 mm (0.001 m) under a 3-kN (3000 N) axial load while keeping the normal stress below 40 MPa (40×106 N/m2), we first calculate the cross-sectional area required using the formula for stress (σ) which is σ = F/A, where F is the force applied and A is the cross-sectional area. Given that the maximum allowable stress σ is 40 MPa, we can reorganize the formula to solve for A, the required cross-sectional area of the rod. This gives us A = F/σ.

Substituting the given values, A = 3000 N / (40×106 N/m2) = 7.5×10-5 m2. To ensure the rod does not exceed the maximum stretch limit when this force is applied, we must also consider the modulus of elasticity (E) for aluminum, which is given as 70 GPa (70×109 N/m2). The formula for elongation (ΔL) under a force is ΔL = (FL)/(AE), where L is the original length of the rod. Given the requirements, the diameter can be calculated from the cross-sectional area (A = πd2/4), where d is the diameter of the rod.

From the area calculated earlier, we can determine the diameter is required to be sufficiently large to maintain stress and elongation within specified limits. Rearranging A = πd2/4 to solve for d, we find d to be approximately 9.8 mm, considering the area necessary to keep stress below 40 MPa while allowing for the specified elongation limit.

Answer:

The dependent variable is MEDV - Median value of owner-occupied homes in $1000's

Explanation:

The median value of the house has to be predicted, based on its properties and neighborhood properties, this can be done by using a linear regression model.

The dependent variable in Machine Learning is the output variable that we want to predict.

Therefore, according to the question given "MEDV" is the dependent variable.

Answer:

Please find attached file for complete answer.

Explanation:

Answer:

Explanation: see attachment below

Answer:

use this to help www.code.org

Explanation:

this helped me alot

The getDenom() and getNum() methods would be sufficient functionality for a user of the Fraction class to determine if two fractions are equivalent, provided that the class also includes methods to compare fractions based on their numerators and denominators.

The getDenom() and getNum() methods are essential for accessing the denominator and numerator of a fraction. With these methods, users can directly compare the numerators and denominators of two fractions to determine if they are equivalent.

By retrieving the numerator and denominator values separately, users can perform mathematical operations or comparisons to check for equivalence without needing additional methods specifically for equivalence testing.

The Complete Question

Explain the importance of the getDenom() and getNum() methods in the Fraction class for determining if two fractions are equivalent.

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

Answer:

[tex]q^.=2.46*10^5W/m^3[/tex]

Explanation:

[tex]Given\\k=2.5W/m\\h_{1} =75(left)\\h_{2} =50(right)\\T_{1} =50^oC\\T_{2} =30^oC[/tex]

so

[tex]T=-\frac{q^.x^2}{2k} +c_{1}x+ c_{2} \\T=T_{1} \\at \\x=-0.04\\T=T_{2} \\at\\x=+0.04[/tex]

[tex]dT/dx=-q^.x/k+c_{1} \\T=T_{max} =300\\at\\x=c_{1} \frac{k}{q^.} (1)[/tex]

[tex]h_{1}(T_{1infinity} -T_{1} )=-k\frac{dT}{dx} |_{x=0.04} (2)\\-k\frac{dT}{dx} |_{x=0.04} =h_{2} (T_{2}-T_{2infinity} (3)[/tex]

[tex]300=-\frac{q^.}{2k} [c_{1} \frac{k}{q} ]^2+c_{1} [c_{1} \frac{k}{q} ]+c_{2} (1)[/tex]

[tex]75[50+\frac{q^2}{2k} (0.04)^2+c_{1} (0.04)-c_{2} ]=-k[\frac{+q^2(0.04)}{2k} ](2)[/tex]

[tex]-k[\frac{-q^.(0.04)}{2k} ]=50[\frac{-q^.(0.04)}{2k} +c_{1} (0.04)+c_{2} -30](3)[/tex]

solving above 3 equations for 3 unknowns c1,c2,q

we get [tex]q^.=2.46*10^5W/m^3[/tex]

Answer:

See step to step explanations for answer.

Explanation:

This is the expression to check if credits is less than 0 assuming variable name is credits:

if(credits<0)

{

credits = 0;

}

Q 4.69:

this is the expression checking the name is Swordfish or not

if(name=="Swordfish")

{

cout<<"We have a match";

}

Q 4.67

This is the expression to check the character assuming that character is ch

if(ch>='1' && ch<='9')

{

cout<<"Digit detected";

}

Q 4.59:

This is the statement to check if the number is between 0 and 500 inclusive . Assuming the variable is n

if(n>=0 && n<=500)

{

cout<<"The number is valid";

}

To address the programming question about conditional expressions, a ternary operator can be used to set the variable 'credits' to 0 if its value is less than 0, and otherwise keep the current value.

The student's question pertains to the use of a conditional expression in a programming context. A conditional expression evaluates a condition and based on its truth value, executes one of two expressions. The statement required should check if the credits variable is less than 0 and if so, assign the value 0 to credits; otherwise, it should leave the value of credits as is.

An example in a generic programming language would be:

This line of code is known as a ternary operator or conditional assignment. It reads as 'Set credits to 0 if credits is less than 0; otherwise, keep the value of credits.'

Answer:

d = 0.868 in

Explanation:

note:

solution is attached due to error in mathematical equation. please find the attachment

This question involves solving for the diameter d at the portion BC of a steel rod under axial load. It involves utilizing Hooke's Law and expressing cross-sectional area in terms of diameter. The final calculation will depend on the length of the portion BC.

The question appears to involve the concept of stress and strain in the area of mechanical engineering. In this problem, we have a steel rod ABC, where end C, experiencing a single axial load P = 15 kips, deflections to 0.05 inches. We are tasked to find the diameter, denoted as d, of portion BC.

To solve this, we could utilize Hooke's Law of elasticity that connects stress, strain, and the Young's modulus (E). The formula is represented as δ = PL/AE. In the equation, P is the load, L is the length, A is the cross-sectional area and E is Young's modulus. We are given P, E, and δ (deflection), leaving the cross-sectional area and length as variables to be found. Since we are looking for the diameter (d), we need to express the area (A) in terms of the diameter using the formula A = πd²/4.

The length L is dependent on the specific conditions of the problem and are not given directly. Depending on the conditions, L might need to be solved differently. First, solve the equation for diameter, afterwards, plug the known values in and evaluate, giving the diameter in inches of the steel rod at portion BC.

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