Hi sir! I am 52 years old mech eng following up your lessons with great pleasure. I would like to kindly remind you that the diameter of pin C should be 0.77 in while pin B has 0.78 in (almost the same) diameter. I presume you underline the importance of double shear pin design in this problem. Thanks and best regards..
I am glad that you caught that the diameter of pin C should be 0.77 because I am a student following his examples and was going crazy thinking I’d done my math wrong.
Should your math on pin C be 7.45 divided by (8*2*0.785) before you take the square root? Did you do the division backwards? I got the diameter at pin C to be .77 inches. Rounded up to 1 inch. Both pins are the same are they not? Thanks!
You are correct. If you listen to him as he is doing the calculations he multiplies the denominator together, than divides it by the numerator (7.45kips). I believe he meant to key his calculator as 7.45kips divided by ANS on his calculator.
I really love your videos. But there's a slight problem. I would prefer you put your ads either at the start or at the end of the video. Since they really ruin the much needed concentration and focus. I would voluntarily watch the ads to give your videos the support they need. Regards
Hello Professor, thank you for making it easy on us. Last step when finding diameter of pin C is 0.77” as I double checked. I think your calculator calculated wrong!!!! Please clarify. Thanks 😊
There are so many ways professors are defining the factor of stress Am soon concluding that the stress given, ( the ultimate or design or alllowable ) watever u call it is always the numerator
Yeah I'm wondering this also. If tau_allowable = 12 kips, with a S.F. of 1.5, tau_actual (as in the actual stress the pin can handle before yielding) should be 18 kips and that's what we design the diameter of the of the pin for... Yes? No?
1newton/1millimeter is equivalent to 1n/1m^3 in other words, 10^3 so, a newton divided by a millimeter is not a megapascal, but a kilopascal. Do the math for 13kn/7853.98m^-3 and you will see that it equals 1655 pascal or 1.65 kPa
Hi Professor, I have a question. Since we are calculating the minimum diameter, aka minimum area A, as to the equation T=F/A, shouldn't we use the Maximum T(T allowable) which is 12ksi, instead of T actual?
This question is asking for the minimum diameter of the pins ~to allow for~ a factor of safety of 1.5. Another way of approaching this problem is to think about T(allowable)/12 kN as the defined value in which the system will fail, and T(actual) as the value we are changing in order to receive a F.O.S of 1.5. Diameter is simply the variable within T(actual) that determines the value of T(actual) itself. When we increase the diameter, the shear stress decreases (diameter is in the denominator) resulting in a lower T(actual) value and a higher F.O.S. Therefore, when we increase the diameter, F.O.S increases as well. This is why the question asks for the ~minimum~ diameter
There is something I am confused about. The allowed shear stress should be smaller than the actual, so shouldn't the actual be allowed times factor of safety which is 12*1.5=18 instead of 12/1.5=8? Other wise people follow the allowed 12 exceeding the actual 8 and gets it damaged.
When looking at the sum of Moments about A, Fab is causing a counter-clockwise rotation. It is positive in that regard. If you are taking the sum of all X forces, then Fab is negative.
corrections:
pin C components: Cx & Cy
resultant at C = 7.45 kips
diameter for pin C: 8=7.45/2[pi(d^2)/4] = 0.77 inches = approx 1 inch
thank you sm! I was confused
Hi sir! I am 52 years old mech eng following up your lessons with great pleasure. I would like to kindly remind you that the diameter of pin C should be 0.77 in while pin B has 0.78 in (almost the same) diameter. I presume you underline the importance of double shear pin design in this problem. Thanks and best regards..
I am glad that you caught that the diameter of pin C should be 0.77 because I am a student following his examples and was going crazy thinking I’d done my math wrong.
Prof. Hanson is the king of the instructors. You might keep on following his lessons..
Thanks for the catch! I was running my numbers over and over again until I saw this comment. Regardless this guy is a lifesaver though.
Exactly my point! I checked out 3 times the calculation, and I found each time, .77 in at B, and .78in at C.
@@emilalopez5353 3 years later and this comment is still blessing me lmao. Just saved me as well
You're such an amazing teacher!
Thank you Prof. Hanson. Your TH-cam channel help me a lot. Pray for me for my upcoming Fe exam! Civil Engineer.
Here reviewing for my FE as well! How did you do?
Pin C should be .77 too otherwise it doesn't make sense, a double shear pin with the same SF should be smaller compare to a single shear.
you have to choose between different diameter pins, .25 inch, .5 inch, .75 inch, 1 inch. as an engineer which one do you choose? OBVIOUSLY the 1 inch
@@Ssm19494 you would choose 1", not 1.5" as he instructed, he divided 12.56/7.45 instead of the other way around.
Should your math on pin C be 7.45 divided by (8*2*0.785) before you take the square root? Did you do the division backwards? I got the diameter at pin C to be .77 inches. Rounded up to 1 inch. Both pins are the same are they not? Thanks!
You are correct. If you listen to him as he is doing the calculations he multiplies the denominator together, than divides it by the numerator (7.45kips). I believe he meant to key his calculator as 7.45kips divided by ANS on his calculator.
@@Jordan-vf4up That should be the correct answer
@@mohammadhamza3937 i got 0.77, he got .78, doesnt really matter when you have to round up to 1 inch for the parameters of the question
i should just skip my lectures and watch yours. These videos are so good
Now I look at things differently. I am that idiot who always overloads things and thinks it's ok. Now I understand why. So informative!
I really love your videos. But there's a slight problem. I would prefer you put your ads either at the start or at the end of the video. Since they really ruin the much needed concentration and focus. I would voluntarily watch the ads to give your videos the support they need.
Regards
Hello Professor, thank you for making it easy on us. Last step when finding diameter of pin C is 0.77” as I double checked. I think your calculator calculated wrong!!!! Please clarify.
Thanks 😊
you have to choose between different diameter pins, .25 inch, .5 inch, .75 inch, 1 inch. as an engineer which one do you choose? OBVIOUSLY the 1 inch
Sir!last diameter of pin C would be .770 instead of 1.298. you placed 7.45 in denumerator instead of placing it in numerator
yea that's what i got too
Same
There are so many ways professors are defining the factor of stress
Am soon concluding that the stress given, ( the ultimate or design or alllowable ) watever u call it is always the numerator
Shouldn’t the factor is safety be FS= sigma(actual)/ sigma(allow) ?
Yeah I'm wondering this also. If tau_allowable = 12 kips, with a S.F. of 1.5, tau_actual (as in the actual stress the pin can handle before yielding) should be 18 kips and that's what we design the diameter of the of the pin for... Yes? No?
The reason the pin diameter error occurred is because he accidentally thought the d^2 term was in the numerator instead of the denominator.
your youtube channel is a life saver !!!
Pin C is 0.77" to nearest 0.25" then, is 1" his answer is correct if you inverse it so it would be (1.298^(-1)) where is = to 0.77"
Fantastic! I'm enjoying it.
Thank you much explaining!
My saviour
Wonderful Lectures ! Thanks.
Jeff, you are amazing.
Thank you 🌹🌺
1newton/1millimeter is equivalent to 1n/1m^3
in other words, 10^3
so, a newton divided by a millimeter is not a megapascal, but a kilopascal.
Do the math for 13kn/7853.98m^-3 and you will see that it equals 1655 pascal or 1.65 kPa
Hi Professor, I have a question. Since we are calculating the minimum diameter, aka minimum area A, as to the equation T=F/A, shouldn't we use the Maximum T(T allowable) which is 12ksi, instead of T actual?
This question is asking for the minimum diameter of the pins ~to allow for~ a factor of safety of 1.5. Another way of approaching this problem is to think about T(allowable)/12 kN as the defined value in which the system will fail, and T(actual) as the value we are changing in order to receive a F.O.S of 1.5. Diameter is simply the variable within T(actual) that determines the value of T(actual) itself. When we increase the diameter, the shear stress decreases (diameter is in the denominator) resulting in a lower T(actual) value and a higher F.O.S. Therefore, when we increase the diameter, F.O.S increases as well. This is why the question asks for the ~minimum~ diameter
@@jaywhaley4688 Ahhh! We're using T(actual)8 instead of 12 just to be conservative aka satisfy the safety factor:D Thank you for the help
how comes the actual stress is less than the allowable one?
There is something I am confused about. The allowed shear stress should be smaller than the actual, so shouldn't the actual be allowed times factor of safety which is 12*1.5=18 instead of 12/1.5=8? Other wise people follow the allowed 12 exceeding the actual 8 and gets it damaged.
Despite these tiny mistakes, you're still the best instructor in this field!
Why arent the x and y reaction forces at c oriented parallel and perpendicular to the pin connection respectively?
The dodge example was soo funny 😂😂
Does the ultimate stress equal the allowable ??
I love you thank you
you are great
If Fab is positive should the force be going to the right?
When looking at the sum of Moments about A, Fab is causing a counter-clockwise rotation. It is positive in that regard. If you are taking the sum of all X forces, then Fab is negative.
Think of positive and negative numbers as verification for your “guess direction.” If the number is positive then you guessed correctly!
If we are designing for complete system why don't we consider area of pin at point A
Its not asked in the qsn being worked
Why do we not use the y component to solve for shear in the pin?
👍👍👍
It would be grateful if you shorten the ads...Thank you
the angle that i got is 63.43 from arctan 4/3 is that wrong? how did you get 60?
it looked like he made it up. how did you get arctan 4/3?
All the letters being confused is confusing to follow