Question

Thin walled tube made of aluminum is loaded with uniformly distributed torque M_0 = 4 kNm/m and concentrated torque M_o^* = 2 kNm at the end Find: M_T(x), tau_max, total angle of rotation phi_A a = 0.2 m, delta = 2 mm, G = 2.5 times 10^4 Mpa

Defleadian Lie anol angle of dleflction af A Thin walleadcbe mace ce aluminiam camiformly distibectecd torgue m-4 kNad concentateal torgsee M'24Nm af the ed Fed: 4)Mr)e) o B)Total auple oaAon s loaakol ith M G=2540* N 4Sm

Answer 1

Solution: In the problem statement a thin-walled triangular shaped tube is given which is under the loading of a uniformly distributed torque and concentrated torqued at the free end. The dimensions and the material properties required are given and we have to find the net moment as a function of distance x from the free end, the maximum shear stress in the whole bar and also the total angle of rotation.

Given:

• 
  Material: aluminum, G 2.5 x 104 MPa
  
• 
  Uniformly distributed torque, Mo 4 kNm/m
  
• 
  Concentrated torque, M 2 kNm
  
• 
  = 0,2 m
  
• 
  2 mm
  

Analysis:

The given bar is subjected to two kinds of torsional loads: (1) concentrated torque at the left end and (2) uniformly distributed torque over the horizontal length of the tube.

Cut the section at a distance x from the free (left) end of the tube. And applying the equilibrium equation in this cut section we can see that there are three torques acting on the element:

1. 
   Uniformly distributed torque, Mo= 4 kNm/m at the left end,
   
2. 
   Concentrated torque, M=2 kNm over the length 'x' of the tube,
   
3. 
   Internal torque, Mr from the right section of the tube.
   ​​​​​​​

Sign convention: We know that generally, an internal torque is positive when its vector points away from the cut section and negative when its vector points towards the section.

Let us assume the internal torque M_x points away in the cut section of the element under consideration. Then applying the equilibrium condition:

M-Mo Mr = 0

Mr Mot M

$\textrm{or } {\color{Teal} \boxed{M_x=4x-2}}$

Now we will calculate the polar moment of inertia for the cross-section of the thin tube. We know that the standard-derived formula for the triangular cross-section of equal sides of length 'a' is given by-

$I_p=\frac{\sqrt{3}}{48}a^4\approx 0.036a^4$

To calculate the polar moment of inertia of the thin tube we will simply subtract the polar moment of inertia of the inner triangular area from that of the outer triangular area. Hence, it is given by:

$I_p=\frac{\sqrt{3}}{48}\left \{a^4-\left ( a-\delta \right )^4 \right \}\approx 0.036\left \{a^4-\left ( a-\delta \right )^4 \right \}$

I, 2.2696698 x 10m

Now we know the function of M_x as well as the polar moment of inertia of the cross-section. To find the maximum shear stress, we will use the torsion formula.

$\boxed{\tau_{max}=\frac{Tr}{I_p}}$

where,

T= applied torque at any cross-section,

r= radial distance of the most distant element in the cross-section from the center

I_p= polar moment of inertia of the cross-section

Here, in this problem, we have calculated above the torque at any section at a distance x given by M_x. To find the value of r we have to calculate the distance of the centroid of the cross-sectional triangle to its vertices. We already know the formula to calculate that. It is given by:

$r=\frac{\textrm{Length of each side}}{\sqrt{3}}=\frac{a}{\sqrt{3}}$

0.2 m0.11547m 3 mar

From the formula given below, we can see that M_x varies from -2 kNm at x=0 to 4 kNm at x=L.   

$M_x=4x-2$

So, the maximum value of the torque is 4 kN-m at the right-hand side of the thin tube. It can be calculated by putting the value of x=L in the above formula.

Now, the maximum shear stress is given by

$\tau_{max}=\frac{M_{x,max}\times r_{max}}{I_p}=\frac{4 kNm \times 0.11547 m}{2.2696698\times 10^{-6}m^4}$

${\color{Teal} \tau_{max}=203,501.06\frac{kN}{m^2}}$

Now we will calculate the total angle of rotation. For a homogeneous material with the constant G, we use the following formula:

$\boxed{\phi =\int_{0}^{L}d\phi=\int_{0}^{L}\frac{T(x)dx}{GI_p(x)}}$

Here, the polar moment of inertia is constant.

Hence, after modifying the above equation we have

$\phi =\frac{1}{GI_p}\int_{0}^{L} M_xdx$

$\phi =\frac{1}{GI_p}\int_{0}^{1.5} (4x-2)dx$

$\phi =\frac{1}{56,741.7456Nm^2}\left | 2x^2-2x \right |_0^{1.5}\textrm{ kN-}m^2$

$\phi =0.02643 \textrm{ radians}$

${\color{Teal} \phi \cong 1.51^{\circ}}$​​​​​​​