A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False

Engineering

1 answer:

ozzi3 years ago

8 0

Answer:

True

Explanation:

For point in xz plane the stress tensor is given by $\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]$

where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components

We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )

thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane

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Nomenclature

T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)