1. Dec 19, 2009

### Fazza3_uae

1. The problem statement, all variables and given/known data

http://img694.imageshack.us/img694/5462/66814773.jpg [Broken]

A new steel bracket was designed to support various loads being hung from the end. The designer would like
to know the largest load that can be hung on the bracket.
What is known:

-Allowable yield stress for the steel is 36 ksi.
-The bracket safety of factor must be at least 2.
-Each arm of the bracket is 6 in long.

2. The Question

What is the maximum load that can be placed on the bracket?

3. The attempt at a solution

First of all i converted all the units into SI units ,, Ksi into N/m2 & in into m

The question gave me the procedure to solve this problem but i am stucking on each step ,, so help is really appreciated friends.

The procedures given :

-Calculate the maximum torsion in the circular bar.
-Calculate the maximum bending stress in the circular bar.
-Determine the stress state at the wall (location of the maximum bending stress and torsional stress)
-Find the principal stresses and using the given factor of safety and the yield stress of steel calculate the

For the first procedure :

Calculate the maximum torsion in the circular bar.

Do they mean the shear stress or it is different .

I used This formula : σmax = T R / Ip

where , σmax is shear stress , T is the torque , R is bar of shaft radius , Ip is moment of inertia.

I found torque by using this formula : T = G * σall

G = Z = Ip/c

σall is given = 248.2e6 N/m2

I found these numbers please if there is any mistake please inform me ,,,

c = diameter / 2 = 0.009525 m

Ip = J = pi/2 * R4 = 1.2929e-8 m4

Then G = Z = J/c = 1.357e-6 m3

Then torque T = G * σall = 336.807 N.m

Finally i found shear stress to be = 248.13e6 N/m2

Which is as same as the σall that is given ,, Is this reasonable ??

Please correct the mistakes if found by you ,,, help me to finish the first step ??

Last edited by a moderator: May 4, 2017
2. Dec 19, 2009

### nvn

Excellent work, so far, Fazza3_uae. Keep going.

3. Dec 19, 2009

### Fazza3_uae

Alright then let's move to the second step .

Calculate the maximum bending stress in the circular bar.

σBending max = M*c / Ip

where M is the bending moment.
c is the maximum vertical distance away from the N.A ( Neutral Axis)
Ip is the moment of inertia.

I found c to be the same as the radius which = 0.009525 m >>> Is it correct ??

Ip is what i've found in the first step >>>> I think this should be right ?

M the bending moment , Here I am confused a little bit , I know that moments and torques are calculated as a force multiplied by distance, but the question here is what force & distance shall i use if i have to use them to solve for M ??

Hmmm now am I going right or missing something ,, ??

Thx nvn for checking my solution

Last edited: Dec 19, 2009
4. Dec 19, 2009

### nvn

The applied force is given in the diagram. Let's call it P. Use the perpendicular distance of P from the wall. Ig is not equal to Ip. Try computing Ig again.

5. Dec 19, 2009

### Fazza3_uae

Ooops i think that I wrote Ig by mistake.Sorry

So now Ip is the same found in the 1st step right ..??

Ok the moment now must = 6 * P In SI units = 0.1524 * P

So

σBending max = 0.1524 * P * 0.009525 / 1.2929e-8

σBending max = 1.1228e5 *P N/m2

Thats it second step done ...

6. Dec 19, 2009

### Fazza3_uae

Now i want to move to step 3.

http://img14.imageshack.us/img14/7660/1copyjm.jpg [Broken]

Determine the stress state at the wall (location of the maximum bending stress and torsional stress)

First of all , Did I locate the stresses correctly with directions , second how to find the components of the maximum bending stress or i don't have to find them ..????

So confused ,,, I am sorry nvn if i ask too much . I really apprectiate ur help.

Last edited by a moderator: May 4, 2017
7. Dec 19, 2009

### Fazza3_uae

The final Step

Find the principal stresses and using the given factor of safety and the yield stress of steel calculate the

I think that ,,

σx = 0
σy = + M*c / Ip ,

M = 1.1228e5 *P , c = 0.009525 m , Ip = 1.2929e-8

$$\tau$$xy = + T*c / Ip ,

T = ??

Now finding principal stresses :

We only need the maximum here right ??

σmax = 0.5 * (σxy) + UnderSquare[ $$\tau$$2xy + 0.5 * (σxy)2 ]

Then after that i use this formula

F.o.S = Ultimate stress / Allowable Stress

Ultimate stress = σMax. = what is found in the last step in terms of P
Allowable Stress = σall. = what is already given in the problem
F.o.S = 2

Is every thing seems good & right ..??

Last edited: Dec 19, 2009
8. Dec 19, 2009

### nvn

Switch the y and z axes in your diagram. Your value of J is correct, but your value of I is wrong. I is not equal to J. Try computing I again.

9. Dec 19, 2009

### Fazza3_uae

http://img268.imageshack.us/img268/1417/1copyr.jpg [Broken]

Ok, I switched the z & y axes in my diagram but why u did that ...??

What about the location of stresses are they gonna stay on there old localtion after switching...?? (( In this case I think that there woulb no more σy or $$\tau$$xy. There would be σz & $$\tau$$zx instead respectively ))

Hmmmm , my value for J I've used in the first step was for the only circular bar.

So do I = the moment of inertia for the whole system ...??

If it is right check my answer . [ The moment of inertia about the x-axis ]

I = (1/3)*(b*h3) + J

where b = 0.1524 m
& h = diameter i think = 0.0127 m

Am i going in the right direction ...??

Last edited by a moderator: May 4, 2017
10. Dec 20, 2009

### nvn

You switched the y and z axes correctly in your diagram. No, do not switch your parameter nomenclature. Keep them as sigmay and tauxy. Your I is still wrong. I = pi*(r^4)/4.

11. Dec 20, 2009

### Fazza3_uae

Ok i will not change anything .

Hmm , Why it is divided by 4

what is the difference b/w this and J

When i calculated J i used R the radius of the shaft not the diameter.
Here you divided by 4 which means to me that r here represents the diameter not the radius... am i right ??

12. Dec 20, 2009

### nvn

No, using your text book, study or look up rectangular area moment of inertia, or rectangular second moment of area, which is called I, for a solid circular cross section. And also study polar area moment of inertia, or polar second moment of area, which is called J, for a solid circular cross section. You got J correct in post 1. But not I.

13. Dec 20, 2009

### pongo38

Stop using formulas for I and J. Instead, go back to first principles with I being the integral of dAy2 and J being ... (you do it). Then you will understand the difference.

14. Dec 20, 2009

### Fazza3_uae

Ok , i have read the textbook and found the difference b/w the Ip Or J and I

I found that the rectangular second moment of area I = pi*(r4)/4

Thanks NvN. Now i have one more question how to find T the torque here :

$$\tau$$xy = = + T*c / I

T = ??

Thanks Pongo38 for the respond i knew the difference now and all the thanks go to NvN.

15. Dec 21, 2009

### nvn

Very good, Fazza3_uae. Torque T = (distance)*P. Fill in the distance (moment arm) here.

16. Dec 24, 2009