SA1 MOHR CIRCILE - Ch 7a - tycnw01.vtc.edu.hktycnw01.vtc.edu.hk/cbe2027/SA1_MOHR_CIRCLE.pdf ·...

CBE2027 Structural Analysis I Chapter 7 – Mohr’s Circle

HD in Civil Engineering Page 7 -1

MOHR’S CIRCLE

A force is easily resolved to components in another coordinate system by a

simple trigonometric calculation. Transformation of stress to another

coordinate system is also a purely mathematical process but is more

involved than the transformation of force.

Stress not only has a magnitude and a direction, but is also associated with

an area over it acts. A transformation of coordinates for stress will change

the area of the orientation plane. This must be taken into account in stress

transformation.

Sometimes we are interested in the state of stress at a particular inclined

plane, e.g. finding the normal and shearing stresses in an inclined glued

splice. More often, we are interested in the maximum intensities of the

normal and shear stresses at a point in the member. For this more general

case, we also need to determine the orientation of the planes where the

maximum normal and shear stresses occur. These are problems to be

addressed in stress transformation.



N = P cos θθθθ V = P sin θθθθ

The normal force N produces positive normal stresses σθ and the shear force

V produces negative shear stress τθ. These stresses can be evaluated by

dividing the forces by the area over which they act. The area A1 of the

inclined section is A/cos θ, in which A is the cross-sectional area.

σθ= N/A1 = (P/A)* cos2 θ = σx cos2 θ

τθ= -V/A1 = -(P/A) sin θ cos θ = -σx sin θ cos θ

in which σx = P/A is the normal stress on a cross-section.



STRESS TRANSORMATION EQUATIONS

An element is subject to the following stresses. We are interested in the

maximum intensities of the normal and shear stresses in the element. For

this case, we need to determine the orientation of the planes where the

maximum normal and shear stresses occur.

Let the thickness of this element ( in the z-direction) be equal to unity.

∑Fn=0

σθds - σx dy cos θ - σy dx sin θ + τx dy sin θ + τy dx cos θ = 0

Given that: dy = ds cos θ, dx = ds sin θ

cos2 θ = (1 + cos 2θ)/2, sin2 θ = (1 - cos 2θ)/2

2 sin θ cos θ = sin 2θ

σθ - σx cos2 θ - σy sin2 θ + 2 τx sin θ cos θ = 0 (Note: τx = τy )

Therefore,

σσσσθθθθ = (σσσσx+ σσσσy)/2 +[(σσσσx - σσσσy)/2] cos 2θθθθ - ττττx sin 2θθθθ (1)

∑Ft = 0

σθ = ?

τθ = ?

n- direction

σy

θ

τy

σx

τx

dy

dx

t- direction



τθds - σx dy sin θ + σy dx cos θ - τx dy cos θ + τy dx sin θ = 0

τθ - σx sin θ cos θ + σy sin θ cos θ - τx cos2 θ + τy sin2 θ = 0

Given that : cos2 θ - sin2 θ = cos 2θ

Therefore,

ττττθθθθ = [(σσσσx - σσσσy)/2] sin 2θθθθ + ττττx cos 2θθθθ (2)

PRINCIPAL STRESSES AND PLANES

By differentiation of the equation (1), we can get the maximum and

minimum normal stresses.

σmax & σmin occur when tan 2θθθθp = 2ττττx/(σσσσy - σσσσx) (3)

Therefore,

σmax = (σσσσx+ σσσσy)/2 + {[(σσσσx - σσσσy)/2]2 + ττττx2}0.5

σmin = (σσσσx+ σσσσy)/2 – {[(σσσσx - σσσσy)/2]2 + ττττx2}0.5

The σmax is called the maximum principal stress while σmin is called the

minimum principal stress.

The planes where the principal stresses occur are called principal planes of

stress.

Equation (3) gives two values of θθθθp that are 900 apart, one for the maximum

stress and another one for the minimum stress.



MAXIMUM SHEAR STRESSES

By differentiation of the equation (2), we can get the maximum and

minimum shear stresses.

The maximum shear stress is equal to

ττττmax = {[(σσσσx - σσσσy)/2]2 + ττττx2}0.5

ττττmin = -{[(σσσσx - σσσσy)/2]2 + ττττx2}0.5

Obviously, the calculations for finding σmax,σmin,τmax and τmin are very

complicate.

By the use of a graphical method (Mohr’s Circle), these values can be

determined very easily.



STRESS ELEMENTS

The following figures show how to draw stress elements for analysis.


HD in Civil Engineering -7



MOHR’S CIRCLE

The Mohr’s circle gives a complete visual picture of the state of stress (or

strain) at a point in a structure and is often easier to use and interpret than a

set of equations.

The symbols for the stresses

The symbols for the stresses have the following meanings:

• A normal stress σσσσ has a subscript that identifies the face on which it acts.

For example, σσσσx acts on the x face of the element.

• A shear stress ττττ has two subscripts: the first indicate the face on which

stress acts, and the second gives the direction of the stress.

For example, ττττxy acts on the x face in the direction of the y axis;

ττττyx acts on the y face in the direction of the x axis.

From consideration of static equilibrium, we note that ττττyx = ττττxy



Sign conventions of Mohr’s Circle

Normal stress σσσσ: Tensile stresses are considered Positive,

Compressive stresses are considered Negative.

Shear Stress ττττ: Positive when it tend to rotate the element Clockwise,

Negative when it tend to rotate the element

Counter-clockwise)

Inclination Angle θ: Positive when measured Counter-clockwise.



Method for Constructing Mohr’s Circle

The scale used for both the σσσσ and τ axes must be the SAME.

• On σσσσ and ττττ axes plot point x have coordinates σσσσx and ττττxy (tend to rotate

element clockwise) and point y having coordinates σσσσy and -ττττxy (tend to

rotate element counterclockwise).

• Draw a line between points x and y. Label the intersection of the line

with the σσσσ axis point O. The coordinates of point O are σσσσ = (σσσσx+ σσσσy)/2

and ττττ = 0.

• With O as the centre, draw Mohr’s circle through points x and y. The

radius of the circle is

R = {[(σσσσx - σσσσy)/2]2 + ττττx2}0.5



Method for Finding Stresses on an Inclined Plane

The method for finding stresses on planes of the element that form angles θ

of with respect to the x and y axes are as follow:

• Rotate the diameter xOy by an angle of 2θ in the SAME DIRECTION as

for the element.

• Locate the points n and t on the circle.

• The coordinates of the point n give the stresses σσσσn and ττττnt.

• The coordinates of the point t give the stresses σσσσt and -ττττtn.

• The values of as the coordinates (σσσσn , ττττnt) and (σσσσt , -ττττtn) may be measured

from the diagram or calculated using geometric relationships.

The important concept to remember is that:

Angles used in Mohr’s circle are double the true angles



Example 1

A stress element is subject to the following stresses.

σσσσx = 0 MPa; σσσσy = 0 MPa ; ττττxy= 40 MPa

1. Draw the initial stress element.

2. Draw the complete Mohr’s circle, labeling critical points.

3. Draw the complete principal stress element.

4. Draw the complete shear stress element.

Solution

The coordinates of point x are σx = 0 MPa and τxy= 40 MPa. The

coordinates of point y are σy = 0 MPa ; τyx= -40 MPa

Center of O of the circle is at:

σ = (σx+ σy)/2 = 0

The radius of the circle:

R = {[(σx - σy)/2]2 + τx2}0.5 = 40 MPa



It is noted that on the principal planes the shear stress is zero.



Example 2

A stress element is subject to the following stresses.

σσσσx = -30 MPa; σσσσy = 20 MPa ; ττττxy= 40 MPa

5. Draw the initial stress element.

6. Draw the complete Mohr’s circle, labeling critical points.

7. Draw the complete principal stress element.

8. Draw the complete shear stress element.

Solution

The coordinates of point x are σx = -30 MPa and τxy= 40 MPa. The

coordinates of point y are σy = 20 MPa ; τyx= -40 MPa


σ = (σx+ σy)/2 = (-30 + 20)/2 = -5 MPa


R = {[(σx - σy)/2]2 + τx2}0.5

= {[(-30 - 20)/2]2 + 402}0.5

= 47.17 MPa



IMPORTANT NOTES

From the Mohr’s Circle, we can note that:

• On the principal planes the shear stress is zero.

• The maximum shear stress occurs where the diameter of the circle is in a

vertical position.

• The magnitude of the maximum shear stress is equal to the radius of the

circle.

• The normal stress σσσσ’ associated with the maximum shear stress is equal

to the average normal stress σσσσave,

σσσσ’ = (σσσσx+ σσσσy)/2

• The planes of maximum shear stress form angles of 450 with the principal

planes.



Example 3

At a certain point in a stressed body, the principal stresses are σx = 80 MPa

and σy = -40 MPa. Determine σ and τ on the planes whose normals are at

=300 and +1200 with the x-axis. Show your results on a sketch of a

differential element.

Solution

The coordinates of point x are σx = 80 MPa and τxy= 0 MPa. The

coordinates of point y are σy = -40 MPa ; τyx= 0 MPa


σ = (σx+ σy)/2 = (80 + -40)/2 = 20 MPa


R = {[(σx - σy)/2]2 + τx2}0.5

= {[(80 + 40)/2]2 + 02}0.5

= 60 MPa



on the 300 face, i.e. at point D

σ = 50 MPa

τ = 52 MPa

on the 1200 face, i.e. at point E

σ’ = -10 MPa

τ’ = -52 MPa



Example 4

A material is subject to two mutually perpendicular direct stresses of +80

MPa and -50 MPa together with a shear stress of 30MPa as shown in the

following figure.

Find

1. the magnitude and nature of the principal stresses;

2. the magnitude of the maximum shear stresses in the plane of the given

stress system;

3. the direction of the planes on which the principal stresses act.

Obtain your answer by means of a Mohr’s circle diagram, and from the

diagram determine the magnitude of the normal stress on a plane inclined at

200 counterclockwise to the plane on which the 50 MPa stress acts.

Solution

The coordinates of point x are σx = -50 MPa and τxy= -30 MPa. The

coordinates of point y are σy = 80 MPa ; τyx= 30 MPa


σ = (σx+ σy)/2 = (80 + -50)/2 = 15 MPa


R = {[(σx - σy)/2]2 + τx2}0.5

= {[(-50 - 80)/2]2 + 302}0.5

= 72 MPa

By measurement from the Mohr’s circle,



σ1 = 87 MPa

σ2 = -57 MPa

τmax= 72 MPa

θ = 155/2 = 77.50

The direct or normal stress on a plane inclined at 200 counterclockwise is

obtained by measuring from point x on the Mohr’s circle through 2x 200 =

400 in the same direction.

This gives

σ = -16 MPa



PRINCIPAL STRSSES AND MAXIMUM SHEAR STRESSES IN BEAM

The normal and shear stresses acting at any point in the cross section of

beam can be obtained from the flexure and shear formulas (σ=My/I and

τ=SAY/It). The normal stress is a maximum at the outer edges of the beam

and equals zero at the neutral axis, whereas the shear stress is zero at the

outer edges and usually is a maximum at the neutral axis.



Example 5

Determine the principal stresses and maximum shear stresses at point C by a

Mohr’s circle construction for the I-section beam as shown in the following

figure.

200 mm

20 mm

20 mm

20 mm

100 mm

C

Solution

The reaction at support A is 40 kN and hence at the required cross-section of

the beam:

M = 40 x 1 = 40 kNm;

V = 40 kN;

I = 100(240)3/12 – 2 x 40 x 2003/12

= 6.19 x 107mm4



The bending stress at point C ,

σc = (40 x 106 x 100)/6.19 x 107

= 64.6 N/mm2

= 64.6 MPa

The shear stress at point C,

τc = [40 x 103 x (100 x 20) x (100+10)] / (I x 20)

= 7.11 N/mm2

= 7.11 MPa

The element C under the above stresses is shown in below:-

By measurement of the above Mohr’s circle,

σ1 = 65.3 MPa

σ2 = -0.77 MPa

τmax= 33 MPa



Problems

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