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Length, Width - Steel Link

 
 
Reply Sat 17 Jan, 2015 04:45 pm
A steel link 3/4 in. in diameter is composed of round steel.

Calculate length of steel used to make the link and weight of link.

Steel weighs .283 lbs. per cu. in.

I have included link to a diagram:
http://www.archive.org/stream/shopmathematics01norriala#page/84/mode/2up

How is the problem solved ?


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Type: Question • Score: 0 • Views: 545 • Replies: 18
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neologist
 
  1  
Reply Sat 17 Jan, 2015 05:18 pm
@Randy Dandy,
Impossible to read diagram on my tablet.
Needed info:
Verify link diameter is ouside.
Verify shape of link as round.
State diameter of round steel

Length of round steel is the average circumference based on inside outside diameter of (round) link.

Treat round steel as if it were a cylinder and calculate volume based on cross sectional area times length.

Good luck.
Randy Dandy
 
  1  
Reply Sat 17 Jan, 2015 06:00 pm
@neologist,
The link diameter is outside.

The link shape is oval - 8 inches across straight parts, 6 inches in length straight parts, radius of curved ends 4 inches.

The diameter is 3/4 inches.

Maybe that will be of help.

Note: I can read the diagram from the link on my pc and tablet. The little magnifying glass with the "plus" sign at the bottom of the page will make the image bigger.


Also, the title of problem should be Length, Weight not Width.
0 Replies
 
neologist
 
  1  
Reply Sat 17 Jan, 2015 07:22 pm
@Randy Dandy,
To begin with,
your fist sentence:
Quote:
A steel link 3/4 in. in diameter is composed of round steel.
Should read:
Quote:
A steel link is composed of round steel 3/4 in. in diameter.
Can you see the difference?
Randy Dandy
 
  1  
Reply Sat 17 Jan, 2015 08:11 pm
@neologist,
Yes, I see the difference.
0 Replies
 
ekename
 
  1  
Reply Sun 18 Jan, 2015 07:41 pm
@Randy Dandy,
Quote:
How is the problem solved ?


The problem is solved by calculating the length of the steel rod (two semi-circles of radius 4 inches and two straight lines of length 6 inches comprise one link). Then multiply that length by the area of the rod thickness (area of a circle with diameter 3/4 of an inch) to obtain the volume of steel. Then calculate the weight based on that volume of steel.

Can you calculate the circumference of two semi-circles of radius 4 inches and add the length of two rods of length 6 inches , as shown in the diagram?
neologist
 
  1  
Reply Sun 18 Jan, 2015 10:15 pm
@ekename,
Since the radius is measured to the center of the 3/4 inch steel, this should be reasonably easy from here.
0 Replies
 
Randy Dandy
 
  1  
Reply Sun 18 Jan, 2015 10:28 pm
@ekename,
Two semi-circles of 4 inches in diameter - 3.1416 *4 *4 = 50.26560

Two straight lines of 6 inches - 12 inches

3.1416 *.75 *.75 = 1.76715


50.26560 + 12 * 1.76715 = 110.03266 * .283 = 31.13924 lbs.

I believe I miscalculated somewhere.
ekename
 
  1  
Reply Sun 18 Jan, 2015 11:11 pm
@Randy Dandy,
Quote:
I believe I miscalculated somewhere.


You must have an answer but not a worked solution if you know you made a mistake. Be more careful selecting the formula to use when answering a question.

Quote:
Two semi-circles of 4 inches in diameter - 3.1416 *4 *4 = 50.26560


The circumference of a circle is 2πr not πr^2.

Quote:
3.1416 *.75 *.75 = 1.76715


The area of the circle is πr^2 not πD^2.


Randy Dandy
 
  1  
Reply Mon 19 Jan, 2015 12:15 am
@ekename,
I don't have an answer.

2 * 3.1416 * 4 = 25.13280
3.1416 * 4 *4 = 50.26560

25.13280 + 50.26560 + 12 = 87.39840

87.39840 * .283 = 24.73375 lbs.

I think I miscalculated again.

Randy Dandy
 
  1  
Reply Mon 19 Jan, 2015 04:25 pm
@Randy Dandy,
Where did I miscalculate ?
neologist
 
  1  
Reply Mon 19 Jan, 2015 04:58 pm
@Randy Dandy,
I'm on my cell phone here and it's difficult to enter text.
Randy, the most important skill you need to learn is how to restate word problems in math terms and vice versa.
I'll get back to this in a few hours to see how you are doing.
Randy Dandy
 
  1  
Reply Tue 20 Jan, 2015 09:17 am
@neologist,
I don't know.
neologist
 
  1  
Reply Tue 20 Jan, 2015 04:05 pm
@Randy Dandy,
You have 2 lengths of steel rod, 6" each = 12"
You have 2 half circles with radius of 4",
essentially 1 circle diameter 8" ; 8 (22/7) = 25 1/7 "

Total length of rod is 37 1/7"
Does that help?
Randy Dandy
 
  1  
Reply Tue 20 Jan, 2015 06:38 pm
@neologist,
Yes, that helps.

So, the weight = 37.142 * .283 = 10.511 = 10.5 lbs.
Randy Dandy
 
  1  
Reply Wed 21 Jan, 2015 09:59 pm
@Randy Dandy,
Correct ?
Randy Dandy
 
  1  
Reply Wed 21 Jan, 2015 10:23 pm
@Randy Dandy,
I was asking if this is correct due to I located a formula for estimating weight of bar stock per foot of length: Multiply diameter of bar by 4, square the product, then divide result by 6 or Weight per foot = (4 x d)^2 / 6. d equals diameter of stock in inches.

For my example - (4 * .75) ^2 = 3 * 3 = 9 / 6 = 1.5

37 1/7 " = 3.095' * 1.5 = 4.6425 lbs.
neologist
 
  1  
Reply Thu 22 Jan, 2015 12:31 pm
@Randy Dandy,
Cross sectional area of stock: pi(3/8)^2 = 0.442 square inches
times length of 37.143 = 16.42 cubic inches
times 0.283 pounds per cubic inch = 4.65 pounds
So, your estimate is correct.
However, it would not work for materials having a different weight per cubic inch.
0 Replies
 
Randy Dandy
 
  1  
Reply Thu 22 Jan, 2015 05:44 pm
Thanks.
0 Replies
 
 

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