Monday, 9 February 2015

STRUCTURAL MECHANICS 2

Tutorial 10:Pressure vessels 
                                                        Solutions
1.      A closed cylindrical steel vessel of radius 2000mm and thickness 10mm has a yield stress of sY = 300MPa.  It contains oil at a pressure of 950 kPa.  Calculate the longitudinal and hoop membrane stresses and deduce the load factor l on membrane yield of the wall.
2.      A cylindrical vessel is being designed with spherical end closures.  The cylinder radius is 3600mm and it is to be made from steel with a yield stress of sY = 260MPa.  The design pressure is 250 kPa and a load factor of l = 1.8 is required against first membrane yield according to von Mises.  Calculate the required thickness of the cylinder, and the required thickness of the spherical ends.  Note that in reality, bending stresses will develop at the junction between the cylinder and sphere, and these will cause first yield at a lower pressure than the membrane theory calculation predicts.
Design pressure  =  250 kPa and load factor  =  1.8.  So take the pressure at failure as  250´1.8 = 450 kPa = 0.450 MPa

Cylinder

Sphere


Note that the spherical ends can be thinner than the cylindrical body.


3.      A large diameter shaft is made as a thin cylindrical shell.  It has a radius of 100mm and a wall thickness 8mm.  It carries a tensile axial load of 930 kN and a torque of 40 kNm.  Calculate the membrane stresses in the cylinder wall, sketch the Mohr circle, and deduce the principal stresses and their orientation to the longitudinal axis.  If the yield stress is 380MPa, find the load factor l against failure according to von Mises criterion.  What should the revised thickness be if the load factor must be increased to l = 2.0.


Sketch Mohr’s circle.  Let longitudinal direction be x and circumferential direction y.
Centre of circle at  C = (sx + sy)/2  =  (185.02 + 0)/2  =  92.51 MPa
Most tensile principal stress  s1 =  C + R  = 92.51 + 122.03 MPa  = 214.54 MPa
Most compressive principal stress  s2 =  C - R  = 92.51 - 122.03 MPa  = -29.52 MPa
Whilst this appears to be a clockwise rotation, the question did not define whether the torque was clockwise or anticlockwise, so either + or – would be OK as an answer.
 4.      A short thin cylindrical aluminium tube with closed ends has a diameter of 600mm and wall thickness 5mm.  It is internally pressurised to a pressure of 2.7MPa.  It carries an axial compressive load of 500 kN and a torque of 180 kNm.  Calculate the membrane stresses in the cylinder wall, sketch the Mohr circle, and deduce the principal stresses and their orientation to the longitudinal axis.  If the yield stress is 230MPa, find the load factor against failure according to von Mises criterion.

Cross-sectional area of tube:  A =  2pRt  =  2p´300´5  =  9424.8 mm2 

This is not a large enough safety margin for most practical purposes

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