CEE 220 - INTRODUCTION TO MECHANICS OF MATERIALS

Tutorial Session 10

Topic: Strain Transformation.  Strain Measurements.  Generalized Hooke's Law.  Failure Theories.

Mohr Circle for Strain

1. Given,

e_x = -8(10^-4)

e_y = 12(10^-4)

g_xy = -15(10^-4)

a) Draw the Mohr circle, and label points X, Y, X_p,Y_p, X_s and Y_s,  Recall that the shear coordinate is e_xy =  g_xy/2.  Compute the radius R, the principal strains e₁ and e₂, and the maximum in-plane shear strain g_max.

R = sqrt[(10)² + (7.5)²](10^-4) = 12.5(10^-4)

e₁ = e_av + R = (2 + 12.5)(10^-4) = 14.5(10^-4)

e₂ = e_av - R = (2 - 12.5)(10^-4) = -10.5(10^-4)

g_max = 2R = 25(10^-4)

b) Show the relative orientation of the axes from point Y_p as origin.  Determine the orientation angle q_p of the major principal axis, as it appears on the Mohr circle.

q_p = tan^-1[-7.5/(10.5-8)] = -71.565 deg.

c) Draw qualitatively the undeformed and deformed shapes of an element ABCD cut along the (x, y) axes, assuming that side AB does not rotate. Do the same for elements cut along the principal axes, and along the axes of maximum shear, rspectively.

2. In a 45^o strain rosette that measures e_a, e_b, and e_c, let the x and y directions be those of e_a and e_c, respectively.  Using the Mohr circle, show that  g_xy/2 = e_b – (e_a + e_c)/2 .

Angle (BC, Bb) = Angle (AC, Ca) because of mutually perpendicular sides.  Thus,

Triangle CBb = Triangle ACa

aA = Cb

or

g_xy/2 = Ob – OC = e_b – (e_a + e_c)/2

Generalized Hooke's Law or  Stress-Strain Relations for Linerly Elastic Isotropic Materials.

3.  a) Write the general stress-strain relations giving (e_x, e_y, e_z) in terms of (s_x, s_y, s_z).

e_x = (s_x -ns_y -ns_z)/E

e_y = (s_y -ns_x -ns_z)/E

e_z = (s_z -ns_x -ns_y)/E

b) In a state of plane stress in the (x, y) plane, s_z = 0.  Obtain in that case (s_x, s_y) in terms of (e_x, e_y).  Obtain also e_z in terms of e_x and e_y.

If s_z = 0,

e_x = (s_x -ns_y)/E

e_y = (s_y -ns_x)/E

Solving for the stresses,

s_x = [E/(1-n²)](e_x + ne_y)

s_y = [E/(1-n²)](e_y + ne_x)

With s_z = 0, we have

e_z = (-ns_x -ns_y)/E =  (-n/E)(s_x + s_y)

s_x + s_y = [E/(1-n²)](1 + n) (e_x + e_y)  = [E/(1-n)](e_x + e_y)

e_z = (-n/E)(s_x + s_y) =  [-n/(1-n)](e_x + e_y)

4.  The strains at a point in a state of plane stress are

e_x = -8(10^-4)

e_y = 12(10^-4)

g_xy = -15(10^-4)

a)  Obtain the principal strains, then the principal stresses using the stress-strain relations of plane stress.  Assume E = 30,000 ksi, and n = 0.3.

e_av = (e_x +e_y)/2 = 2(10^-4)

R = sqrt{[(e_x -e_y)/2]² + (g_xy/2)²}= 12.5(10^-4)

e₁ = e_av + R = 14.5(10^-4)

e₂ = e_av - R = -10.5(10^-4)

s₁ = [E/(1-n²)](e₁ + ne₂) = 37.42 ksi

s₂ = [E/(1-n²)](e₂ + ne₁) = -20.27 ksi

b)  Outline another procedure for determining the principal stresses.

Obtain the (x, y) stresses by the stress-strain relations, then determine the principal stresses.

5. Why are the principal axes of strain coincident with the principal axes of stress in an isotropic material?

Because both are characterized  by zero shear.

6. A thin square plate fits tightly on its bottom face and on its four edges in a square cavity in an infinitely rigid material.  A uniform pressure p is applied on the upper face of the plate.  Assuming all contacts with the cavity are smooth, determine the reaction stresses on the edges of the plate.  Let the (x, y) axes be in the plane of the plate.  Assume  material properties E and n, and constant stresses.

Hint: What can you say about the stresses on the upper face of the plate?  What can you say about the shear stresses on the surfaces of contact with the sides of the cavity?  What can you say about the (x, y) strains?

The applied pressure imposes s_z = -p, and zero shear stresses on the top face.  The sides of the rigid cavity impose zero displacements, and thus e_x = e_y = 0.  Smooth contacts impose zero shear stresses. The conditions e_x = e_y = 0 yield

s_x – ns_y – ns_z  = 0

s_y – ns_x – ns_z = 0

Solve for s_x and s_y, and substitute s_z = -p:

s_x = s_y = ns_z/(1- n) = -np/(1- n)

Failure Criteria

7.  Let the maximum bending stress in a beam in planar bending be s_max.

a) If the shear stress on an area element in the cross-section at the point where s_max occurs is zero, is it consistent with the Tresca criterion and with the Mises-Hencky criterion that yield would start when s_max = s_Y?  Why or why not?

Yes, it is consistent with both criteria, because the state of stress is uniaxial, and both criteria describe the onset of yield in that state by the condition s_max = s_Y.

b) Give one example of a cross section where the flexural shear stress is not zero at a point where s_max occurs.

In an I section, the maximum bending stress occurs at the top and bottom of the flanges.  The shear stress in a flange, parallel to the flange, is not zero.

8.  a) Draw the Tresca hexagon and the Mises-Hencky ellipse in the plane (s₁, s₂), and define key points of these curves in terms of the yield stress s_Y of the material.

b) In a cylindrical pressure vessel, the hoop stress s₁ is twice the axial stress s₂.  Show graphically the points representing the onset of yield according to the two criteria.  For what value of s₁ does yield occur according to the Mises-Hencky criterion?

c) Show graphically the points representing the onset of yield for a spherical pressure vessel.

d) Show graphically the points representing the onset of yield for a bar of circular cross section in torsion.