M5.1 Torsion

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Mechanics
of materials
Torsion
Plan de rescate 2018
1. Las calificaciones del 1er parcial se reportarán como:
• 60% examen
• 40% quizzes y prototipo de examen
2. Habrá un parcial extra que incluirá sólo los temas de carga axial
y torsión el día 5 de octubre (después de la sesión de repaso)
3. De los tres parciales tomaré los dos más altos para calcular la
calificación final
Torsion
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Torsion
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TORSION
Linear distribution of stress
Torsion – shear relationship:
Polar moment of inertia J
The resistance opposed by a shaft or beam to being distorted by
torsion, as a function of its shape
For a solid shaft:
For a tubular shaft:
=
=
−
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EXAMPLE 1
The pipe shown in Fig. 5–12a has an inner diameter of 80 mm and an outer
diameter of 100 mm. If its end is tightened against the support at A using a torque
wrench at B, determine the shear stress developed in the material at the inner and
outer walls along the central portion of the pipe when the 80-N forces are applied
to the wrench.
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EXAMPLE 1 (cont)
Solutions
The only unknown at the section is the internal torque T
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 M y  0;
800.3  800.2  T   0
T  40 N  m
The polar moment of inertia for the pipe’s cross-sectional area is
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J

0.05  0.04   5.79610  m
4
2
4
6
4
For any point lying on the outside
lsurface of the pipe,
  c0  0.05 m
0 
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Tc0
400.05

 0.345 MPa (Ans)
J
5.796 10 6


Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 1 (cont)
Solutions
And for any point located on the inside surface,
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i 
Tci
400.04

 0.276 MPa (Ans)
J
5.796 10 6


The resultant internal torque is equal but opposite.
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EXAMPLE
5.2
EXAMPLE
5.2
(CONTINU
ED)
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EXAMPLE
5.2
(CONTINU
ED)
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Copyright ©2014 Pearson Education, All Rights Reserved
Copyright ©2014 Pearson Education, All Rights Reserved
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EXAMPLE
5.1
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EXAMPLE
5.1
(CONTINU
ED)
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