Howt to calculate shear stress?

[janisr]

I have a cross-section of a beam like this one http://imageshack.us/photo/my-images/810/img0129xf.jpg/
And I have to calculate the shear stress τ.
I have these things already calculated:
Xc=8cm; Yc=6.9cm
Moment of Inertia Ix=2946.7 cm^4
Moment of resistance Wmin=323.8 cm^3; Wmax=427.1 cm^3
σ+=23.41 KPa; σ-=30.88 KPa
How do I get the shear stress? And where do I look for information?

Thanks!

 

[nvn]

janisr: You would look in, and study, a mechanics of materials textbook, under the topic "shear flow" (but not in the chapter on torsion). You must list relevant equations yourself, and show your work. And then someone might check your math. There is a template for homework questions. Also, the unit symbol for kilopascal is kPa, not KPa. Uppercase K means kelvin. Also, always leave a space between a numeric value and its following unit symbol. E.g., 6.9 cm, not 6.9cm. However, I would use N, mm, MPa.

 

[janisr]

Thanks for reply! :)

I have question for things I have already done.
If I have Moment of resistance Wmin=323.8 cm^3; Wmax=427.1 cm^3 then I get σ+= (Mmax/Wmaxx) and σ-= (Mmax/Wmin) right?
Professor gave M=10 kNm. And this was the number I puted into those formulas.

 

[janisr]

So the whole problem is in the picture. Professor just draw the cross-section of a beam and wrote down the things I had to find + he wrote M=10 kNm.
I think I managed to calculate sigma_plus and sigma_minus. As for sigma_minus - it is the tension in the part of a beam cross-section that is pressed/squeezed (Sorry for my english). And if we asume that the point load is on y-axis (going downwards) then the pressed part is above the x axis, accordingly stretched part is below x axis. Knowing that, I can calculate moment of resistance in the stretched part and in the pressed part. W_stretched= I/6.9 and W_pressed= I/9.1
And these I use to find sigma_plus= M_max/W_stretched and sigma_minus= M_max/W_pressed
Please tell me if I am doing this completely wrong.
And now I have to find τ, I read in the book that you have to use this τ=(Q_max*Sx)/(Ix*b)
where Sx is static moment (how do you call it in english?). But the shear force Q is not given and I do not know what to do.

 

[nvn]

janisr: Nice work. Yes, posts 3 and 4 are correct, if your Ix value is correct (which I did not check). Sx is called statistical moment of area, or sometimes first moment of area.

With no applied force given, no type of applied force given (such as point load, or uniformly-distributed load), no free-body diagram of the beam, and no boundary conditions given, then you cannot know the shear force, Qmax. You could assume Qmax is anything from zero to perhaps 0.062 N (?), if your current M value is correct. Are you saying there is no free-body diagram (FBD) showing the beam, the applied force, and the constraints? Is this a steel beam? It has very little applied moment.

If you are sure your extremely small M value is correct, then you could perhaps assume the beam is a horizontal cantilever, of length L = 160 mm, having a tip point load P, pointing upward (or downward), if you wish. Therefore, Pmax = Mmax/L = (10.00 N*mm)/(160 mm) = 0.0625 N. And Qmax = Pmax. (You cannot realistically make L shorter than 160 mm, due to Saint-Venant's principle.)

 

[janisr]

All that is given is in the picture in post #4. And with this I can not calculate shear stress τ, am I right?
If so, maybe you are right and I am supposed to assume Qmax or I have to draw a beam and put on some kind of loads, but then again the Mmax will change. The book we use in our studies is useless in this situation.

 

[nvn]

janisr: Right. With what is given in posts 1, 3, and 4, you cannot compute Q nor Qmax. Perhaps see the revisions and new (last) paragraph I added in post 5.

 

[janisr]

M is correct because it was given. Maybe It is small because it is just an exercise homework for my exam. And I think I'll go with L=160 mm, as you suggested. No diagrams was given, nor material was specified. Tomorrow I hope I will meet my teacher. Thanks for your time, it was helpful. :)