Plain Strain

UPIK: Materials Technology III

PLANE STRAIN

1) A cylindrical storage tank used to transport gas under pressure has an inside diameter of 600mm and a wall thickness of 20mm. Strain gauges attached to the surface of the tank in transverse and longitudinal directions indicate strains of 250µ and 60µ respectively. If the modulus of rigidity Ԍ for the tank material is Ԍ = 80ԌPa, determine

a) the gauge pressure inside the tankb) the principal stresses and the maximum shearing stress in the wall of the tank

SOLUTION:

a) the gauge pressure inside the tank

We note that the given strain are the principal strains at the surface of the tank. Plotting the corresponding points A and B, we draw Mohr's circle for strain. The maximum in-plane shearing strain is equal to the diameter of the circle:

Mutumba George Instructor


N.B:

From Hooke's law for shearing stress and strain we have:

= (80 x 109Pa)(195 x 10-6) = 15.60MPa

Recall that

Where σ2 is the longitudinal stress and σ1 is the hoop stress.

Therefore

p = 4.16MPa

b) the principal stresses and the maximum shearing stress

Recall that for thin walled cylindrical vessels, σ1 = 2σ2 and draw Mohr's circle for stress:



σ2 = 2τmax (in plane) = 2 x 15.60 = 31.2MPa

σ1 = 2σ2 = 2 x 31.2 = 62.4MPa

The maximum shearing stress is equal to the radius of the bigger circle of diameter OA and corresponds to a rotation of 45˚ about a longitudinal axis.

τmax = ⅟2σ1 = σ2 = 31.2MPa

2) Using a 60˚ rosette, the following strains have been determined at point Q on a vertical surface of a steel machine base:

ϵ1 = 40µ, ϵ2 = 980µ ϵ3 = 330µ

Using the coordinate axes shown to determine at point Q

a) the strain components ϵx , ϵy, ᴕxy

b) the principal strains

c) the maximum shearing strain, use = 0.29



SOLUTION:

a) the strain components ϵx , ϵy, ᴕxy

For the co-ordinate axes shown Ɵ1 = 0˚, Ɵ2 = 60˚, Ɵ3 = 120˚

The equations to be used are:

ϵ1 = ϵxcos2Ɵ1 + ϵy sin2Ɵ1 + ᴕ xysin Ɵ1cos Ɵ1

ϵ2 = ϵxcos2Ɵ2 + ϵy sin2Ɵ2 + ᴕxysin Ɵ2cos Ɵ2

ϵ3 = ϵxcos2Ɵ3 + ϵy sin2Ɵ3 + ᴕxysin Ɵ3cos Ɵ3

Substitute in the above equations:

ϵ1 = ϵx x 1 + ϵy x 0 + ᴕxy x 0 x 1

ϵ2 = ϵx(0.5)2 + ϵy (0.866)2 + ᴕxy(0.866) x (0.5)

ϵ3 = ϵx(-0.5)2 + ϵy (0.866)2 + ᴕxy(0.866) x (-0.5)

Solving the above equations gives:

ϵx = ϵ1 ϵy = ⅓(2ϵ2 + 2ϵ3 x 1 - ϵ1)

Substituting the given values for ϵ1, ϵ2 and ϵ3:

ϵx = 40µ,

ϵy = ⅓(2 x 980 + 2 x 330 - 40) = 860µ

ᴕxy = (980-330)/0.866 = 750µ



The strains are indicated on the element below:

b) the principal strains

We note that the side of the element associated with ϵx rotates counterclockwise, thus we plot X below the horizontal axis ie X(40, -375). We then plot Y(860, +375) and draw Mohr's circle:



ϵave = ⅟2(860µ + 40µ) = 450µ

R = √[(375µ)2 + (410µ)2] = 556µ

tan2Ɵp = 375µ/410µ

Points A and B correspond to the principal strains and we have

ϵa = ϵave - R = 450µ - 556µ

= -106µ

ϵa = ϵave - R = 450µ - 556µ = ϵave + R = 450µ + 556µ

= +1006µ

Since σz is zero on the surface, the equation for plain strain below can be used to the principal

strain ϵc:

= -368µ



c) the maximum shearing strain

We plot point C and draw Mohr's circle through B and C to get point D':


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Plain Strain