Solve Mechanics: Maximum Load Needed? Urgent Help

[Fazza3_uae]

Mechanics Question : Maximum Load needed ?? Urgent Please Help

Homework Statement

http://img694.imageshack.us/img694/5462/66814773.jpg

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?

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
allowable Load.


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 : σ_max = T R / I_p

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

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

G = Z = I_p/c

σ_all 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

I_p = J = pi/2 * R⁴ = 1.2929e^-8 m⁴

Then G = Z = J/c = 1.357e^-6 m³

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

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

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

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



Thx in advance for all.

 

[nvn]

Excellent work, so far, Fazza3_uae. Keep going.

 

[Fazza3_uae]

Alright then let's move to the second step .

Calculate the maximum bending stress in the circular bar.


σ_{Bending max} = M*c / I_p

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

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

I_p 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

 

[nvn]

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

 

[Fazza3_uae]

Ooops i think that I wrote I_g by mistake.Sorry

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

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

So

σ_{Bending max} = 0.1524 * P * 0.009525 / 1.2929e^-8

σ_{Bending max} = 1.1228e⁵ *P N/m²

Thats it second step done ...

 

[Fazza3_uae]

Now i want to move to step 3.

http://img14.imageshack.us/img14/7660/1copyjm.jpg

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.

 

[Fazza3_uae]

The final Step

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

I think that ,,

σ_x = 0
σ_y = + M*c / I_p ,

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

$$\tau$$_xy = + T*c / I_p ,

T = ??

Now finding principal stresses :

We only need the maximum here right ??


σ_max = 0.5 * (σ_x+σ_y) + UnderSquare[ $$\tau$$²xy + 0.5 * (σ_x-σ_y)² ]

Then after that i use this formula

F.o.S = Ultimate stress / Allowable Stress

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

Is every thing seems good & right ..??

 

[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.

 

[Fazza3_uae]

http://img268.imageshack.us/img268/1417/1copyr.jpg


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

What about the location of stresses are they going to stay on there old localtion after switching...?? (( In this case I think that there woulb no more σ_y or $$\tau$$_xy. There would be σ_z & $$\tau$$_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 ...??

 

[nvn]

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

 

[Fazza3_uae]

You switched the y and z axes correctly in your diagram. No, do not switch your parameter nomenclature. Keep them as sigma_y and tau_xy.

Ok i will not change anything .

Your I is still wrong. I = pi*(r^4)/4.

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 ??

 

[nvn]

No, using your textbook, 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.

 

[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.

 

[Fazza3_uae]

Ok , i have read the textbook and found the difference b/w the I_p 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.

 

[nvn]

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

 

[Fazza3_uae]

yup , i have done it. thanks a lot NvN.

all credit goes to u . thanks a ton.