Stress and Strain

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Andrew_Risk

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1) Calculate the tensile stress in a 33.000 mm diameter rod subjected to a pull of 49.000 kN.

Answer 1 = MPa (3 decimal places) 57.298???

2) Consider that the rod was originally 1.000 metres long, and is stretched 1.000 mm by the pulling force. Calculate the strain produced in the rod.

Answer 2 = (6 decimal places) 0.001000???

A steel bar has a rectangular cross section measuring 1.000 cm by 2.300 cm and is 8.700 m in length. A load of 159000.000 N causes the bar to extend 1.200 mm. Calculate the stress and strain produced.

Answer 1 = kPa (3 decimal places)

Answer 2 = (6 decimal places)


The ultimate strength of a round steel rod is 550.000 MPa. If a factor of safety of 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 7.200 cm?

Answer = kN (Round to 3 decimal places)
What is the factor of safety of a steel hanger having an ultimate strength of 630.000 MPa and supporting a load of 61000.000 N. The steel hanger in question has a cross sectional area of 5.400 cm2.

Answer = (Round to 3 decimal places)
I practice these all the time, I am just not getting it, I have class tomorrow, so maybe things will go better, but after 4 days of practice questions, I always end up getting them wrong. Thanks for any help. My teacher is retiring at the end of the year, so he really doesn't care, I've paid for a tutor, he showed up twice and always has an excuse on why he cant meet me for assistance, What I don't understand is the usage of 10^-6 or 10^-4, I am lost. Again thanks for any help.
 
Andrew_Risk said:
I am just not getting it.... after 4 days of practice questions, I always end up getting them wrong.
Click to expand...
What have you done? What were your steps? What answers did you get? What are the "right" (or at least "expected") answers?

Andrew_Risk said:
What I don't understand is the usage of 10^-6 or 10^-4....
Click to expand...
I'm not seeing either of these in what you've posted...?
 
Andrew_Risk said:
1) Calculate the tensile stress in a 33.000 mm diameter rod subjected to a pull of 49.000 kN.

Answer 1 = MPa (3 decimal places) 57.298???

stress = Force/area = 49000/[π/4 * 332 ) N/mm2 = 57.2899 MPa = 57.3 MPa

2) Consider that the rod was originally 1.000 metres long, and is stretched 1.000 mm by the pulling force. Calculate the strain produced in the rod.

Answer 2 = (6 decimal places) 0.001000???

strain = (final_length - original_length)/(original_length) = (deformation)/(original_length) = 1/1000 = 0.001000

A steel bar has a rectangular cross section measuring 1.000 cm by 2.300 cm and is 8.700 m in length. A load of 159000.000 N causes the bar to extend 1.200 mm. Calculate the stress and strain produced.

Answer 1 = kPa (3 decimal places) Follow the method of problem 1.

Answer 2 = (6 decimal places) Follow the method of problem 2.



The ultimate strength of a round steel rod is 550.000 MPa. If a factor of safety of 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 7.200 cm?

Load = stress * Area

Allowable stress = ultimate strength/ (Factor of safety) = 550/3 MPa = 550/3 N/mm2

Allowable load = Allowable stress * Area = 550/3 * π/4 * (72)2 N ............ finish it

Answer = kN (Round to 3 decimal places)
What is the factor of safety of a steel hanger having an ultimate strength of 630.000 MPa and supporting a load of 61000.000 N. The steel hanger in question has a cross sectional area of 5.400 cm2.

Area = 5.4 cm2 = 5.4 * 102 mm2

First find the applied stress on the hanger = 61000/(5.4 * 100) N/mm2.......................(1)

Factor safety = (ultimate strength)/(applied stress) ............... finish it

Answer = (Round to 3 decimal places)
I practice these all the time, I am just not getting it, I have class tomorrow, so maybe things will go better, but after 4 days of practice questions, I always end up getting them wrong. Thanks for any help. My teacher is retiring at the end of the year, so he really doesn't care, I've paid for a tutor, he showed up twice and always has an excuse on why he cant meet me for assistance, What I don't understand is the usage of 10^-6 or 10^-4, I am lost. Again thanks for any help.
Click to expand...

If you still have problem - come back and show your work. We can help better that way.
 
Is this right?

1. A steel bar has a rectangular cross section measuring 1.000 cm by 2.300 cm and is 8.700 m in length. A load of 159000.000 N causes the bar to extend 1.200 mm. Calculate the stress and strain produced.

Answer 1 = kPa (3 decimal places)

Answer 2 = (6 decimal places)

covert cm to meters

0.01*0.023=0.00023
for stress: 159000/0.00023=691304347.835kPa rounded to 3 decimal places

for strain:1.200/8.700=0.1379310344827586? I never rounded this to 6 decimal places.


2. The ultimate strength of a round steel rod is 600.000 MPa. If a factor of safety of 5.500 is required, what is the maximum permissible load for the rod if it has a diameter of 6.300 cm?

(600.000 x 10^6 N/m^2) * pi * (0.06300 m)^2 / 5.500 = 1360.252 kN This was how one of the other students solved his, but I don't understand where he gets 10^6 from? any ideas?
Thanks for any help.
 
Have no idea on what to do for these questions:
3. The ultimate strength of a round steel rod is 550.000 MPa. If a factor of safety of 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 7.200 cm?

4. What is the factor of safety of a steel hanger having an ultimate strength of 630.000 MPa and supporting a load of 61000.000 N. The steel hanger in question has a cross sectional area of 5.400 cm2.

5. Calculate the tensile stress in a 33.000 mm diameter rod subjected to a pull of 49.000 kN.


6a. Calculate the tensile stress in a 33.000 mm diameter rod subjected to a pull of 49.000 kN.

Answer 1 = MPa (3 decimal places)

6b. Consider that the rod was originally 1.000 metres long, and is stretched 1.000 mm by the pulling force. Calculate the strain produced in the rod.

Answer 2 = (6 decimal places)
 
Andrew_Risk said:
1. A steel bar has a rectangular cross section measuring 1.000 cm by 2.300 cm and is 8.700 m in length. A load of 159000.000 N causes the bar to extend 1.200 mm. Calculate the stress and strain produced.

Answer 1 = kPa (3 decimal places)

Answer 2 = (6 decimal places)

covert cm to meters

0.01*0.023=0.00023
for stress: 159000/0.00023=691304347.835kPa rounded to 3 decimal places

for strain:1.200/8.700=0.1379310344827586? I never rounded this to 6 decimal places.


2. The ultimate strength of a round steel rod is 600.000 MPa. If a factor of safety of 5.500 is required, what is the maximum permissible load for the rod if it has a diameter of 6.300 cm?

(600.000 x 10^6 N/m^2) * pi * (0.06300 m)^2 / 5.500 = 1360.252 kN This was how one of the other students solved his, but I don't understand where he gets 10^6 from? any ideas?
Thanks for any help.
Click to expand...

for the stress check your calculator. I get 691304347.826 Pa (not kPa) and this is 691304.347 kPa.

for the strain you need to keep units consistent. You have (1.2 cm)/(8.7 m). This isn't correct, you need to convert the numerator units to meters (or the denominators to cm, it doesn't really matter).

for #2 I don't see where your confusion lies. You have the rod diameter so you can calculate the area of a cross section. The load divided by this area gives you the stress. The maximum allowable stress is the "ultimate strength" of the rod divided by the safety factor. The 10^6 is just because the strength of the rod is given in MPa rather than just Pa. (M is for mega or million, i.e. 10^6). So just chug the numbers and come up with your max allowable load.
 
Andrew_Risk said:
Have no idea on what to do for these questions:
3. The ultimate strength of a round steel rod is 550.000 MPa. If a factor of safety of 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 7.200 cm?

4. What is the factor of safety of a steel hanger having an ultimate strength of 630.000 MPa and supporting a load of 61000.000 N. The steel hanger in question has a cross sectional area of 5.400 cm2.

5. Calculate the tensile stress in a 33.000 mm diameter rod subjected to a pull of 49.000 kN.


6a. Calculate the tensile stress in a 33.000 mm diameter rod subjected to a pull of 49.000 kN.

Answer 1 = MPa (3 decimal places)

6b. Consider that the rod was originally 1.000 metres long, and is stretched 1.000 mm by the pulling force. Calculate the strain produced in the rod.

Answer 2 = (6 decimal places)
Click to expand...
same as

http://www.freemathhelp.com/forum/threads/84777-Stress-and-Strain?p=348313#post348313
 
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