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Hollow rectangular beam  rigidity and max torque. 
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#1
Apr2111, 01:04 PM

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Revision that is turning into a learning from scratch session. Some help would be welcome! (I have laso put this on the materials section of engineering).
 I have a hollow rectangular beam. 0.150m by 0.110m with uniform thickness walls at 0.003m. It is 3m long.  It has a modulus of rigidity (G) of 27 GigaPascals.  It has a max permitted shear stress of 200 MegaPascals. What is the torsional rigidity? What is max permitted torque (with a safety margin of 2.0)? What would be the twist angle for the 3m length? Torsional rigidity = G x second area of inertia. I have G so need 1xx. 1xx = Each side at ((0.003x0.110^3)/12) = 3.33 E07 Top & base each at (using parallel axis theory) ((0.003 x 0.144^3)/12) + (0.003 x 0.144 x 0.062^2) = 2.41 E06 Add 4 sides = 5.48 E06 Torsional rigidity = 27 E09 x 5.48 E06 = 148,000 (units???) Is this right and what are the units? As for the other 2 parts – I am lost. Any pointers? 


#2
Apr2111, 01:11 PM

P: 4,512

As for your last two questions, they are related. Is there a constraint other than structural failing, such as the yield strength that determines the max allowable twist angle? Euler buckling of the walls is also a possibility.



#3
Apr2111, 01:16 PM

P: 8

Phrak  thanks for replying. There are no other contraints in this question.



#4
Apr2111, 01:25 PM

P: 4,512

Hollow rectangular beam  rigidity and max torque.
Well, that's it then. Find the twist angle that hit's 200 MegaPascals stress for any given element. This is one possible limit. The other limit is determined by buckling, dependent upon the modulus of rigidity.



#5
Apr2111, 02:09 PM

P: 8

Phrak, Thanks. What equation would I use to find the twist angle that uses the information that I already have?
I think it may be twist angle = (torque x length) / (torsional rigidity). I have torsional rigidity (above  if it is correct?) and I have length (3m) but I am not sure about torque and how would I apply a safety factor of 2.0? I know I need to use the max permitted shear stress of 200MegaPascal but I am not sure how. Thanks. 


#6
Apr2111, 03:58 PM

P: 4,512

Well, I haven't seen your equations. I'm familiar with circular cross sections, where the stress is simple to calculateit's pure compression and tension cotangent with a helical curve running at 45 degree around the axis. What do your stresses look like with a rectangular cross section?



#7
Apr2111, 04:16 PM

P: 8

The approach to the question is supposed to be very basic in order to help us with revision (it needs to be for my level of understanding) so I doubt it will get into 3d .
I think that to get the twist angle I need torque (as above) but to get torque I need Torque = 2 x area x shear flow(q).......... but to get the shear flow ..... i need Torque... I am not sure how to get out of that circle with what I have got. 


#8
Apr2111, 06:00 PM

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I don't know all the details behind this assignment, but the torsional stress analysis of hollow, noncircular cross sections is completely different from analyzing circular shafts.
For instance, given the dimensions of your cross section and the thickness of the sides, this beam could be treated as a thinwalled beam. A torque applied to the beam sets up shear stresses in the walls, such that tau = T / (2At), where T = torque, t = wall thickness, and A = total area enclosed by the mid line of the sides [including the hollow portion]. For a complete description of the details, you must consult special texts on advanced strength of materials. 


#9
Apr2211, 03:53 AM

P: 4,512

but I'm still betting on Euler buckling before material failure. 


#10
Apr2211, 04:46 AM

P: 8

Phrak and Steam King  thanks. Whilst it would be interesting to see if Eular Buckling occurred sooner than material failure I am confident that I am expected to find the material failure figures.
This brings me to Steam King's equation "tau = T / (2At), where T = torque, t = wall thickness, and A = total area enclosed by the mid line of the sides [including the hollow portion]". Does this mean that the tau is the maximum permissible shear stress (the 200MegaPascals that I have been given) and that I can get torque by rearranging the equation to be T= tau x (2At). This will give me torque to enter into the equation twist angle = (torque x length) / (torsional rigidity). The numbers should work out to be: Torque = 200E6 (tau) x 2 x 0.147x0.107 (enclosed area) x 0.003 (wall thickness) = 18,800 (units ?Nm?) However I am supposed to have a 2.0 safety factor so I divide this by 2 = 9,400 I can then use this to work out the twist angle. Twist angle = (9,400 (torque) x 3 (length)) / 148,000 (torsional rigidity) = .19 (is this radians? If it is then this would be 11 degrees. I think I may have gone wrong somewhere? Any suggestions? 


#11
Apr2211, 11:00 AM

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The J calculation for thinwalled closed noncircular sections is not related to finding the usual second moments of area typically encountered in bending stress calculations.
For a closed thinwalled noncircular section, J = 4tA^2/S, where t = thickness [constant thruout the section], A is the cross sectional area [same area used to determine tau], and S is the total perimeter length around the cross section at the mid line of the sides. The angle of twist, phi = tau * S / (2*A*G), where tau, S, and A are defined as above. G is the modulus of rigidity. These relations work only for noncircular sections having constant thickness. For more general sections, you must consult advance strength of materials texts. 


#12
Apr2211, 11:36 AM

P: 8

Steam King, thanks.
1. So I should not have used second moment of inertia or indeed parallel axis theorem because they do not apply to torques only to bending moments. 2. I should have used torsional rigidity = GJ. I have G so all I need to do is to calculate J by using your equation. In this case: 4 x 0.003 (wall thickness) x (0.147x0.107)^2 (enclosed area) / 0.496 (the perimeter). So J = 5.99E6. Torsional rigidity = 5.99E6 x 27E9 = 161730 (units?) 3. For the angle of twist I use phi = 200E6 (tau) x 0.496 (the perimeter) / 2 x (0.147x0.107) (enclosed area) x 27E09 (G) = 0.117 radians or 6.69degrees. Am I now in the right ball park? Again – thanks. 


#13
Apr2211, 04:07 PM

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You could check that J = Ix + Iy



#14
Apr2211, 06:56 PM

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Except that J = Ix + Iy only for circular cross sections.



#15
Apr2411, 04:09 PM

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Very well said by SteamKing. Only one minor mistake. You need to multiply the numerator of the phi equation by L.
10^3: Regarding your questions in post 12: (1) No, second moment of area applies also to torsion, for closed, circular cross sections, as foreshadowed by SteamKing. (2 and 3) Your perimeter, s, is slightly incorrect. See the definition by SteamKing for parameter s, and try again on items 2 and 3. Regarding units, you can figure out units yourself by including units in your arithmetic and algebra. Try again. Also, to include the safety factor, divide maximum permitted shear stress by the safety factor. By the way, always leave a space between a numeric value and its following unit symbol. E.g., 0.150 m, not 0.150m. See the international standard for writing units (ISO 310). Also, torque is N*m, not Nm. And the correct spelling is megapascal (MPa), gigapascal (GPa), newton (N). 


#16
Apr2811, 04:08 PM

P: 8

Steam King and nvk. Many thanks. It really is appreciated. Third time lucky! Here we go….
1. I calculate torsional rigidity = GJ. 2. J = 4tA^2/S. 4 x 0.003 (wall thickness) x (0.147x0.107)^2 (enclosed area) / 0.508 (the perimeter .147+.147+.107+.107). So J = 5.84E6 m^4 (correct units?). 3. Torsional rigidity = GJ = 5.84E6 x 27E9 = 157,791 n*m^2 (correct units?). 4. Max shear stress (200 Mpa) / safety factor (2) = 100 MPa. 5. Max permitted torque with a safety factor: = 100E6 (tau) x 2 x 0.147x0.107 (enclosed area) x 0.003 (wall thickness) = 9,437 N*m. 6. For the angle of twist I use: twist angle = (torque x length) / (torsional rigidity). = [9437 N*m (max permitted torque) * 3 m (length)] / 157,791 n*m^2 (torsional rigidity) = 0.18 radians or 10.28 degrees. Concerns: A, Should I be using second moment of inertia? GJ or GIxx? I get J to be 5.84E6 m^4 and Ixx to be 5.48 E06 m^4. They are pretty close but which one is correct? B. I have worked through all of my calculations using the units and have ended up with those shown above. Have I now got this right? Nvk – thanks for your tips and I agree that precision is important. C. I have 2 ways to calculate the angle of twist: twist angle = (torque x length) / (torsional rigidity) (giving 0.18 radians) and phi = 100 E6 (tau) x 0.508 (the perimeter) / 2 x (0.147x0.107) (enclosed area) x 27E09 (G) = 0.059 radians (I presume that this would be per metre so I need * 3m to get 0.18 radians). It is good news that they agree, however, which is the more appropriate method or are they in fact identical and I cannot see it? Once again  THANK YOU! 


#17
Apr2811, 09:54 PM

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10^3: All your answers appear correct, except you misspelled some unit symbols. Always use the correct capitalization of unit symbols. E.g., N*m, not n*m. And MPa, not Mpa. Also, numbers less than 1 must always have a zero before the decimal point. E.g., 0.147, not .147. See the links I posted.
You should use J, not Ixx. Therefore, you used the correct value for J. Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three (or four) significant digits. You can use either twist angle formula you prefer. I prefer your second formula. 


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