me 227 lecture 3

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Name: Dr. Muhammad Rabiul IslamRank: MajorPosition: Assistant Professor/ Instructor Class BMobile: 01915489917Email: [email protected]

Course Teacher

ME 227 LECTURE (PART – 3)

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CHAPTER 7Analysis of Stress and Strain

The stresses in a beam are given by the flexure and shear formulas

(=My/I and = VQ/Ib). The stresses calculated from these formulas

act on cross sections of the members, but larger stresses may occur

on inclined sections. Therefore, we will begin our analysis of

stresses and strains by discussing methods for finding the normal

and shear stresses acting on inclined sections cut through a member.

We have already derived expressions for the normal and shear stresses acting on inclined sections in uniaxial stress.

In the case of uniaxial stress, we found that the maximum shear stresses occur on planes inclined at 45° to the axis, whereas the maximum normal stresses occur on the cross sections.

Analysis of Stress and Strain

Stresses are not vectors. This fact can sometimes be confusing,

because we customarily represent stresses by arrows just as we

represent force vectors by arrows.

Although the arrows used to represent stresses have magnitude and

direction, they are not vectors because they do not combine

according to the parallelogram law of addition. Instead, stresses are

much more complex quantities than are vectors, and in mathematics

they are called tensors.

Other tensor quantities in mechanics are strains and

moments of inertia.

Analysis of Stress and Strain

When the material is in plane stress in the xy plane, only the x and y faces of the element are subjected to stresses, and all stresses act parallel to the x and y axes.

A normal stress has a subscript that identifies the face on which the stress acts. The sign convention for normal stresses is the familiar one, namely, tension is positive and compression is negative.

A shear stress has two subscripts—the first subscript denotes the face on which the stress acts, and the second gives the direction on that face.A shear stress is positive when the directions associated with its subscripts are plus-plus or minus-minus; the stress is negative when the directions are plus-minus or minus-plus.

from equilibrium of the element:

Stresses on Inclined Sections

Two Dimensional View of Plane-Stress

The shear stresses acting on all four side faces of an element in plane stress are known if we determine the shear stress acting on any one of those faces.

Wedge-Shaped Stress Element

Area of the left-hand side face : A0 ;

Normal and shear forces : xA0 and

xyA0;

Area of the bottom face :A0tan;

Area of the inclined face : A0 sec.Normal and shear forces acting on those faces have the magnitudes and directions shown in Fig.

By summing forces in the x1 direction:

What if =00 or 900??

Stresses on Inclined Sections

Transformation Equations for Plane Stress

Some Trigonometric Identities:

If these substitutions are made, the equations of Stresses on Inclined Sections become:

These equations are called the

transformation equations for plane

stress.

By substituting +90° for .

Summing the expressions for x1 and y1

Special Case – I (Uniaxial Stress)(All stresses acting on the element are zero except for the normal

stress x)

Special Case – II (Pure Shear)(x = 0 and y = 0)

Special Case – III (Biaxial Stress)(Element is subjected to normal stresses in both the x and y

directions butwithout any shear stresses)

A plane-stress condition exists at a point on the surface of a loaded structure, where the stresses have the magnitudes and directions shown on the stress element of Fig. at right. Determine the stresses acting on an element that is oriented at a clockwise angle of 15° with respect to the original element.

Results can be verified by substituting =75° and by the equation x1 + y1 = x + y.

Principal Stresses

the principal stresses occur on mutually perpendicular planes.

Since:

Principal Stresses:

Principal Angles

Shear Stresses on the Principal Planes

Putting x1y1 = 0 and solving:

The shear stresses are zero on the principal planes.

Maximum Shear Stresses

Comparison Shows: Planes of maximum shear stress

occur at 45° to the principal planes

OR

the maximum shear stress is equal to one-half the difference of the principal stresses

Review of Class Test - 02

Calculation for Shear Force & Bending Moment Diagram

Calculate the value of S.F. by intensity of load and:

Join the line of forces by:

Calculate the value of B.M. by intensity of moment and:

Join the line of B.M. by:

Equation of B.M. may require to understand the nature of line.

Review of Class Test - 03

It seems you know, but you cannot write!!

No one could answer the both questions properly. Some derived the equation properly but couldn’t answer the math. Other solve the math partially but couldn’t derive the equation. I do not think the time was insufficient.

None could solve the math completely!!

My Assessment:

Understanding is not the main problem. You cannot keep your brain cool in exam and there is lack of preparation. Please write down all the derivations and solve each type of problem by yourselves following the steps.

Very few are very much weak. For them, “Please be careful and try more. Otherwise you will be in great problem”.

Review of the Relations before the MOHR’s Circle

Principal Stresses:

Principal Angles:

The shear stresses are zero on the principal planes.

Maximum Shear Stresses:

Transformation equations for plane stress:

Planes of maximum shear stress occur at 45° to the principal planes

MOHR’S CIRCLE FOR PLANE STRESS

(1)2 + (2)2 (2) ;

(1)

Considering:

which is the equation of a circle in standard algebraic form. The coordinates are x1 and x1y1, the radius is R, and the center of the circle has coordinates x1

= aver and x1y1 = 0.

;

Equations of Mohr’s Circle

Why MOHR’s Circle?? If stresses x , y, and xy acting on the x and y planes of an element in plane stress are known;


With the circle drawn, we can determine the stresses x1, y1, and x1y1 acting on an inclined element;

Also the principal stresses and maximum shear stresses can be obtained from the circle.

x1

x1y1

Construction of Mohr’s Circle

(x , y, and xy are known)

A ( = 0)

x

xy

B ( = 90)y

-xy

C


Calculate the stresses x1, y1, and x1y1 for inclination

x1

x1y1

A ( = 0)

x

xy

B ( = 90)y

-xy

2D ( = )

C

x1

x1y1

R Cos

R Si

n

Finally the stress-transformation equations for x1 and x1y1 can be obtained.

D

( = +90)

y1


Principal Stresses, Principal Angles and Maximum Shear Stresses

x1

x1y1

A ( = 0)

x

xy

B ( = 90)y

2P1

CP1P2

1

2

O

2P1 = (2 + )

S1

S2 = R

Solving the Problems

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CHAPTER 9Deflections of Beams

In Chapter 5 we used the curvature of the bent beam to

determine the normal strains and stresses in the beam.

However, we did not develop a method for finding the

deflection curve itself. In this chapter, we will determine

the equation of the deflection curve and also find

deflections at specific points along the axis of the beam.

A problem oriented very important chapter

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DIFFERENTIAL EQUATIONS OF THE DEFLECTION CURVE

Beams with Small Angles of Rotation

;

This equation can be integrated in each particular case to find the deflection v,

provided the bending moment M and flexural rigidity EI are known as functions of x.

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OF BEAMS WITH SMALL ANGLES OF ROTATIONReminder of the sign conventions

(1) The x and y axes are positive to the right and upward, respectively;

(2) The deflection v is positive upward;

(3) The slope dv/dx and angle of rotation are positive when counterclockwise with respect to the positive x axis;

(4) The curvature k is positive when the beam is bent concave upward;

(5) The bending moment M is positive when it produces compression in the upper part of the beam.

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OF BEAMS WITH SMALL ANGLES OF ROTATIONNonprismatic

Beams

Prismatic Beams

Using this notation:

These equations are referred as the bending-moment equation, theshear-force equation, and the load equation, respectivelyThe general procedure of solving problems consists of integrating the equations and then evaluating the constants of integration from boundary and other conditions pertaining to the beam.

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Solving the Differential Equations of the DeflectionCurve and Obtain Deflections of Beams

For each region of the beam, we substitute the expression for M into the differential equation and integrate to obtain the slope v.

Each integration of the expression for M produces one constant of integration. Next, we integrate each slope equation to obtain the corresponding deflection v. Again, each integration produces a new constant. Thus, there are two constants of integration for each region of the beam. These constants are evaluated from known conditions pertaining to the slopes and deflections. The conditions fall into three categories: (1) boundary conditions, (2) continuity conditions, and (3) symmetry conditions.

Expression for the bending moments can be obtained from free-body diagrams and equations of equilibrium

Boundary Conditions

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Solving the Differential Equations of the DeflectionCurve and Obtain Deflections of Beams

Continuity Conditions

Symmetry condition: For instance, if a simple beam supports

a uniform load throughout its length, we know in advance

that the slope of the deflection curve at the midpoint must be

zero.

me 227 lecture 3

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