ENG204 – Strength of MaterialsLecture 10, Sections 7.1-7.4, 7.9- Recall that the general state of

stress at a given point is given by 6 components: σx, σy, σz, τxy, τyz, τzx.

- As previously indicated, this state of stress can be represented by a different set of components if the coordinates axes is rotated.

9:05 Dr. Ammar T. Al-Sayegh 0

State of Plane Stress- Our discussion of the transformation of stress will deal

mainly with plane stress—the situation in which two of the faces of cubic element are free of any stress. i.e., σz = τzx = τzy = 0, so the remaining stresses are σx, σy, and τxy.

9:05 Dr. Ammar T. Al-Sayegh 1

Transformation of Plane Stress- Assuming a state of plane stress, we need to determine

the stress components σx’, σy’, and τx’y’ after the element has been rotated through an angle θ about the z axis.

9:05 Dr. Ammar T. Al-Sayegh 2

Transformation of Plane Stress- Consider a prismatic element with faces perpendicular to

the x, y, and x’ axes.

9:05 Dr. Ammar T. Al-Sayegh 3

θθτθσθσσ cossin2sincos0 22'' xyyxxxF ++=→=∑

)sin(coscossin)(0 22''' θθτθθσστ −+−−=→=∑ xyyxyxyF

- Recall that

- and

- The equations can be rearranged to

− σy’ can be obtained from σy’ by replacing θ with θ + 90.

9:05 Dr. Ammar T. Al-Sayegh 4

θθθ cossin22sin =

θτθσσσσ

σ 2sin2cos22' xy

yxyxx +

−+

+=

θθθ 22 sincos2cos −=

22cos1cos2 θθ +

=2

2cos1sin2 θθ −=

θτθσσ

τ 2sin2sin2'' xy

yxyx +

−−=

θτθσσσσ

σ 2sin2cos22' xy

yxyxy −

−−

+=

- Note 1: note that the sum of normal stresses exerted on a cubic element of material is independent of the orientation of that element. i.e.

- Note 2: If you square both equations obtained for the transformation of stresses and add them, then set

you get the parametric equation of a circle with a radius R, centered at point C of abscissa σave in the form

9:05 Dr. Ammar T. Al-Sayegh 5

yxyx σσσσ +=+ ''

2yx

ave

σσσ

+= 2

2

2 xyyxR τ

σσ+

−=

( ) 22''

2' Ryxavex =+− τσσ

- Note that points A and B correspond to the maximum and minimum normal stresses. At these values, shearing stress is zero. To get these values, we set τx’y’ = 0 in

to get

which defines 2 θp which are 90 apart. These are the principle planes of stress.

9:05 Dr. Ammar T. Al-Sayegh 6

θτθσσ

τ 2sin2sin2'' xy

yxyx +

−−=

yx

xyp σσ

τθ

−=

22tan

9:05 Dr. Ammar T. Al-Sayegh 7

- It can be observed from the circle

substituting for σave and R,

- Referring again to the circle, it can be seen that τmax equals R

and it’s located at

Rave +=σσmax Rave −=σσmin

22

minmax, 22 xyyxyx τ

σσσσσ +

−±

+=

22

max 2 xyyx τ

σστ +

−=

xy

yxs τ

σσθ

22tan

−−=

Mohr’s Circle for Plane Stress

- The circle used to determine stresses was introduced by Otto Mohr, thus called Mohr’s Circle, and can be used instead of formulas to determine solve plane stress problems.

- To draw Mohr’s circle, first plot points X (σx , -τxy) and Y (σy , τxy), then connect X and Y with a line defining the diameter and center of the circle.

9:05 Dr. Ammar T. Al-Sayegh 8

Angle Directions in Mohr’s Circle

9:05 Dr. Ammar T. Al-Sayegh 9

Stresses in Cylindrical Thin-Walled Pressure Members

9:05 Dr. Ammar T. Al-Sayegh 10

- Consider a cylindrical vessel on inner radius r and wall thickness t. select a small element on the wall with sides parallel and perpendicular to the axis of the cylinder.

- Due to the axisymmetry, there will be no shearing stress. Thus, σ1(hoop) and σ2 (longitudinal) are principle stresses, given by

- While maximum shearing stress istpr

=1σ tpr22 =σ

tpr22max ==στ

Stresses in Spherical Thin-Walled Pressure Members

9:05 Dr. Ammar T. Al-Sayegh 11

- Consider a spherical vessel on inner radius r and wall thickness t containing fluid of pressure p.

- Due to the symmetry, stresses on all faces will be equal. Thus, σ1 = σ2and are principle stresses, given by

- Maximum shearing stress occurs at 45o and given by

tpr221 ==σσ

tpr42

11max == στ

Problem 7.164- Two steel plates of uniform cross section 10 x 80 mm are

welded together as shown. Knowing that centric 100 kNforces are applied to the welded plates and that β = 25o, determine (a) the in-plane shearing stress parallel to the weld, (b) the normal stress perpendicular to the weld..

9:05 Dr. Ammar T. Al-Sayegh 12

Problem 7.158- The grain of a wooden

member forms and angle 15o with the vertical edge. For the state of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the grain.

9:05 Dr. Ammar T. Al-Sayegh 13

Problem 7.122- A torque of magnitude T

= 12 kN.m is applied to the end of a tank containing compressed air under pressure to 8 MPa. Knowing that the tank has a 180-mm inner diameter and a 12-mm wall thickness, determine the maximum normal and shearing stresses.

9:05 Dr. Ammar T. Al-Sayegh 14