Stress Analysis Question - Please Help!!!!!!!!

[leeave96]

Stress Analysis Question - Please Help!!!!!!!!

OK, I have confused myself again!

I have a part that is made of steel. It is a cantilevered tab. The length of the cantilevered tab is 1.125". The height of the tab is .25" and the width of the tab is 1".

I am trying to solve for the max amount of weight I can rest on the tip of this tab.

Soooo:

Stress (S) = Mc/I

M = lbs x L

L = 1.125

h = .25", therefore c = .125"

b = 1"

I = (b x h^3)/12 => (1 x .25^3)/12 => 0.0013in^4

Rearaanging the stress formula and substituting in the values for M, I get:

S = lbs x L x c / I

Solving for lbs:

lbs = S x I / c x L => (S x 0.0013) / (.125 x 1.125)

My question is - for stress (S), do I use the modulus of elasticity (which for this problem is 30 x 10^6 psi) or yield strength in shear (which for this problem is 36 x 10^3 psi)?

With MoE, I get 277,333 lbs.

With yield, I get 322.8 lbs.

The yield answer seems more like it is correct to me.

Again, I have gotton myself confused and my second question is when solving for an allowable stress in such a problem as shown above, when does one use MoE or yield?

Thanks much!
Bill

 

[jbeard]

You are correct that you should be using yield strength for the calculation you have here.

The modulus of elasticity would come into play if you wanted deflection, hooke's law:
S=Ee
or
stress = (modulus of elasticity) * strain

Or in your case the deflection at the end or the beam
d=(PL^3)/(3EI)
where d is deflection, P is the load, L is length, E is modulus, and I is moment of inertia.

Hope that helps,
Jamie

 

[Willbur]

Bill, you really should use the yield strength in tension (not shear) since the bending formula you're using actually calculates the tension/compression in the outer surfaces of the part. As in all engineering design calculations,

:!: Don't forget the safety factor :!:

 

[Cameraman]

Willbur is right all the way around:

. . . use the Sy in tension (because you are examining bending, not shear . . .

. . . and be sure to include a safety factor! This latter point is critical because it is what permits you to make the simplifying assumptions (such as c=t/2) that make your analyses easy.

Regards and good luck,
Greg

 

[McGuyver]

Re:

I concur with Wilbur and cameraman. You will want to use the tensile yield strength of steel in this equation. And by all means, use a safety factor, otherwise, all you will be calculating is the load at which the beam will experience peramanent plastic deformation due to bending. Good luck!