The pressurized pipe shown below is closed from both ends and is subjected to an
ID: 1843380 • Letter: T
Question
The pressurized pipe shown below is closed from both ends and is subjected to an internal pressure of p = 5 MPa and a transverse load of F = 1 kN in the -z direction. The pipe can be thought of a thin-walled vessel with an inside diameter of d_i = 2R = 200 mm and a thickness of t = 5 mm. Find the stresses acting at the stress element A on the pipe outer surface as shown above. Sketch and label Mohr's circle corresponding to the state of stress at A. Determine the principal normal and extreme shear stresses at A (sigma_1, 2 and tau_1, 2). Find the principal normal stress and extreme shear stress directions or angles (Phi_P and Phi_S) and then sketch the principal normal and extreme shear stress elements at A. What is the maximum shear stress at A (tau_max)?Explanation / Answer
a. Balancing forces and couples in x, y and z axis. We get F= 1kN, M= 1kN*3m=3kN.m and T= 2kN.m.
The moment of inertia for the shell is I= pi*R^3*t = 1.57*10^-5 m^4. From this calculate Zp =Ip/y. Y being radius of the shell = 100 mm.
Stresses acting on element A:
1. Bending stress (sigma1)= M*y/I = 3000*100/(1.57*10^-5) = 19.1 MPa
2. Shear due to torsion (tau1) = T/Zp = 2000/ (3.14*10^-4) = 6.37MPa
3. Hoop stress due to pressure (sigma2) = pR/t = 5*100/5 =100MPa
4. Longitudnal stress due to end closure of pipe (sigma 3) = 50 MPa
On element A: Stress in x direction = Sigma x = Sigma 1 + Sigma 3 = 69.1 MPa
Stressin y-direction= Sigma y = SIgma 2 = 100 MPa
Shear stress = tau = 6.37MPa
b.
c. Maximum stress is Sigma = [(Sigma x + Sigma y)/2] + sqrt [((sigma x-sigma y)/2)^2+ (tau)^2]
= [(69.2+100)/2]+[{(69.2-100)/2}^2 + (6.37)^2] = 101.26MPa
d. Direction of principal normal stress = arctan {2*Tau/ (sigma x - sigma y)} = arctan {2*6.37/ (69.1-100)}=22.4
e. Maximum shear stress = sqrt [{(sigma x-sigma y)/2}^2 + tau^2] = 16.71 MPa
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