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Strength Of Materials Miscellaneous

Strength Of Materials

  1. A thin cylinder of 100 mm internal diameter and 5 mm thickness is subjected to an internal pressure of 10 MPa and a torque of 2000 Nm. Calculate the magnitudes of the principal stresses.
    1. 109.75 & 40.25 MPa
    2. 19.75 & 40.25 MPa
    3. 101.75 & 41.25 MPa
    4. 19.75 & 44.25 MPa
    5. None of these
Correct Option: A

di = o.1 m, do = di + 2t
t = 0.005 m, do = 0.11 m
T = 2000 Nm
P = 10 MPa

σc =
pdi
=
10 × 106 × 0.1
= 50 MPa
4t4 × 0.005

σc =
pdi
=
10 × 106 × 0.1
= 100 MPa
2t2 × 0.005

T
=
τ
Jr

τ =
Tr
J


= 200 ×
0.11
2
π
(0.114 - 0.14)
32

σ1 , 2 =
σx + σy
± √
σx - σy
² + τ²xy
22

=
50 + 100
± √
50 - 100
² + 24.142
22

= 75 ± 34.75
= 109.75 & 40.25 MPa



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