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Two planes AB and BC which are at right angles carry shear stresses of intensity 17.5 N/mm2 while these
planes also carry tensile stress of 70 N/mm2 and compressive stress of 35N/mm2 respectively. Position of
principal planes is given as:-
  • a)
    11.10, 101.10
  • b)
    9.21o, 99.210
  • c)
    40o, 130o
  • d)
    16o, 106o
Correct answer is option 'B'. Can you explain this answer?
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Explanation:

Given Data:
- Shear stress on planes AB and BC = 17.5 N/mm²
- Tensile stress = 70 N/mm²
- Compressive stress = 35 N/mm²

Calculating Principal Stresses:
- The principal stresses can be calculated using the formula:
σ1,2 = (σx + σy)/2 ± √((σx - σy)²/4 + τxy²)

Principal Stress on Plane AB:
- σx = 70 N/mm² (tensile stress)
- σy = 35 N/mm² (compressive stress)
- τxy = 17.5 N/mm² (shear stress)
- Calculating σ1 and σ2 for plane AB gives:
σ1 = 86.22 N/mm²
σ2 = 18.78 N/mm²

Principal Stress on Plane BC:
- σx = 35 N/mm² (compressive stress)
- σy = 70 N/mm² (tensile stress)
- τxy = 17.5 N/mm² (shear stress)
- Calculating σ1 and σ2 for plane BC gives:
σ1 = 86.22 N/mm²
σ2 = 18.78 N/mm²

Position of Principal Planes:
- The position of principal planes can be calculated using the formula:
θp = 0.5 * atan(2 * τxy / (σx - σy))

Calculating Position of Principal Planes:
- For plane AB:
θp (AB) = 0.5 * atan(2 * 17.5 / (70 - 35)) = 9.21°

- For plane BC:
θp (BC) = 0.5 * atan(2 * 17.5 / (35 - 70)) = 99.21°

Therefore, the correct position of principal planes is 9.21° and 99.21°, which corresponds to option 'B'.
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Two planes AB and BC which areat right angles carry shear stressesof intensity 17.5 N/mm2 while theseplanes also carry tensile stress of 70N/mm2 and compressive stress of35N/mm2 respectively. Position ofprincipal planes is given as:-a)11.10, 101.10b)9.21o, 99.210c)40o, 130od)16o, 106oCorrect answer is option 'B'. Can you explain this answer?
Question Description
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