1. The state of stress (MPa) at a point is given by σx =13.78 = , σy = 26.2, σz = 41.34, τxy = 27.56, τyz = 6.89 and τxz = 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:

α = 50° 70’ ; β= 60° ; γ = ?

l = Cos α = 0.62705; m = Cos β = 0.50.

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

which implies γ = Cos-1 0.597 = 53° 23’

S = 56.609 and

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

2. At a point in a material, the state of stress in MPa is given by the components σx = 12.8, σy

= 27, σz= 51.3, τxy = 23.4, τyz = -6.24 and τxz = 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 α = 0.66913; m = Cos β = 0.3255.

We know that l2 +m2+n2 = 1 and hence n = Cosγ = 0.6680 ° 5’ 3’’

σ = =48.79

S2=2532.9 and S = 50.328

Directions of shear stress are

3. A stress resultant of 140 Mpa makes angles 43ᵒ,75ᵒ and 50ᵒ 53’ with x, y and z axes. Determine the normal and shear stresses on the oblique plane whose normal makes respective angles 67ᵒ 13’, 30ᵒ and 71ᵒ 34’ with these axes.

Solution:

S=140;

Mpa

Mpa

Mpa

4. A resultant stress of 170 Mpa is inclined at 23 and 72 with x and y axes respectively. The

resultant stress acts on a plane with normal cosines 0.656, 0.259 relative to x and y respectively. Determine the normal and shear stresses on this plane. Given the components shear stresses are

Find the normal stresses with invariant J1 = 926.4 Mpa.

Solution:

S = 170 Mpa;

Also Sz= 40.68 Mpa

.

5. Find the invariants and principal stresses for the following stress components in Mpa: =9.68

what will

be the principal stresses be? Show that principal stress directions are orthogonal.

Solution:

The invariants are given by

J3 = 9.68 (14.32 x 17.28 – 7.382) – 2.44 (2.44 x 17.28 – 10.21 x 7.28) + 10.21 (2.44 x 7.38 – 10.21 x 14.32) = 640.15

The principal stresses are given by :

Which gives = 28.38 Mpa ; = 10.8 Mpa ; = 2.08Mpa;

ii) When , the stress determinant becomes

The principal stress directions are given by

A = 71.25 ; B = 0 and C = 0; ;

l1=0.10, m1 = 0 and n1 = 0

l2=1 , m2 = 0 and n2 = 0

And for l3=1 , m3 = 0 and n3 = 0

Hence the principal stress directions are orthogonal.

6. Given that the partial state of stress at a point is

=20 The stress invariant J1 = 50 Mpa .

Determine the remaining normal stresses, principal stresses and maximum shear stress.

J1 =

The stress tensor is and J1 = 50 Mpa; J2 = -1100 and J3 = -7000

Solving the above cubic equation ,