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Strength Of Materials Miscellaneous

Strength Of Materials

  1. The state of stress at a point on an element is shown in figure (a). The same state of stress is shown in another coordinate system in figure (b).

    The components (τxx, τyy, τxy,) are given by
    1. p
      , -
      p
      , 0
      22
    2. (0,0, p)
    3. p, - p,
      p
      2
    4. 0, 0,
      p
      2
Correct Option: B


σθ =
σx + σy
+
σx - σy
cos2θ + τxysine2θ
22

Here θ = - 45
σθ = Txy
σx = p
σy = - p
σθ =Txy =
P + P
+
P - P
cos90° = 0
22

When θ = + 45
σθ =
P - P
+
P + P
cos90°
22

When θ = 45° Tθ = Txy
Tθ =
σx - σy
sin2θ - Txy cos2θ
2

Tθ = Txy =
P + P
sin90° = P
2

∴ Txx, Tyy, Txy = 0, 0, p



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