The state of stress (MPa) at a point is given by x =13.78 = , y = 26.2, z = 41.34, xy = 27.56, yz = 6.89 and xz = 10.34. Determine the stress resultants and the normal and shear stresses for a plane whose normal makes angles of 50 70 and 60 with

The state of stress (MPa) at a point is given by x =13.78 = , y = 26.2, z = 41.34, xy = 27.56, yz = 6.89 and xz = 10.34. Determine the stress resultants and the normal and shear stresses for a plane whose normal makes angles of 50 70 and 60 with the directions of x and y respectively. Solution:

= 50 70 ;

= 60 ;

= ?

l = Cos = 0.62705;

m = Cos = 0.50.

We know that l2 +m2+n2 = 1 and hence n = Cos = 0.597

which implies = Cos-1 0.597 = 53 23

S = 56.609 and

EMBED Equation.KSEE3 \* MERGEFORMAT

Stress Resultants Sx = 28.59 ; Sy = 34.49; Sz = 34.608 and ;

At a point in a material, the state of stress in MPa is given by the components x = 12.8, y = 27, z= 51.3, xy = 23.4, yz = -6.24 and xz = 11 Determine the normal and shear stresses on a plane whose normal makes angles of 48 and 71 to the x and y axes respectively. Also find the direction of the shear stress relative to x and y.Solution:

l = Cos = 0.66913;

m = Cos = 0.3255.

We know that l2 +m2+n2 = 1 and hence n = Cos = 0.6680 5 3

= =48.79

S2=2532.9 and S = 50.328

Directions of shear stress are

A stress resultant of 140 MPa makes angles of 43,75 and 5053' with the x and y axes. Determine the normal and shear stresses on an oblique plane whose normal makes respective angles of 67 13'',30and 7134 with these axes.Solution:

S=140

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